discrete_mathematics_with_applications/chapter_5/exercises.md

Exercise Set 5.1

Page 296

Write the first four terms of the sequences defined by the formulas 1-6.

1. a_k = \dfrac{k}{10 + k}, for every integer k \geq 1.

 a_1 = \frac{1}{10 + 1} = \frac{1}{11} 

 a_2 = \frac{2}{10 + 2} = \frac{2}{12} 

 a_3 = \frac{3}{10 + 3} = \frac{3}{13} 

 a_4 = \frac{4}{10 + 4} = \frac{4}{14} 

2. b_j = \dfrac{5 - j}{5 + j}, for every integer j \geq 1.

 b_1 = \dfrac{5 - 1}{5 + 1} = \frac{4}{6} 

 b_2 = \dfrac{5 - 2}{5 + 2} = \frac{3}{7} 

 b_3 = \dfrac{5 - 3}{5 + 3} = \frac{2}{8} 

 b_4 = \dfrac{5 - 4}{5 + 4} = \frac{1}{9} 

3. c_i = \dfrac{(-1)^i}{3^i}, for every integer i \geq 0.

 c_0 = \dfrac{(-1)^0}{3^0} = \frac{1}{1} 

 c_1 = \dfrac{(-1)^1}{3^1} = \frac{-1}{3} 

 c_2 = \dfrac{(-1)^2}{3^2} = \frac{1}{9} 

 c_3 = \dfrac{(-1)^3}{3^3} = \frac{-1}{27} 

4. d_m = 1 + \left(\dfrac{1}{2}\right)^m for every integer m \geq 0.

 d_0 = 1 + \left(\dfrac{1}{2}\right)^0 = 1 

 d_1 = 1 + \left(\dfrac{1}{2}\right)^1 = \frac{1}{2} 

 d_2 = 1 + \left(\dfrac{1}{2}\right)^2 = \frac{1}{4} 

 d_3 = 1 + \left(\dfrac{1}{2}\right)^3 = \frac{1}{8} 

5. e_n = \left\lfloor \dfrac{n}{2} \right\rfloor \cdot 2, for every integer n \geq 0.

 e_0 = \left\lfloor \dfrac{0}{2} \right\rfloor \cdot 2 = 0 

 e_1 = \left\lfloor \dfrac{1}{2} \right\rfloor \cdot 2 = 0 

 e_2 = \left\lfloor \dfrac{2}{2} \right\rfloor \cdot 2 = 2 

 e_3 = \left\lfloor \dfrac{3}{2} \right\rfloor \cdot 2 = 2 

6. f_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 4, for every integer n \geq 1.

 f_1 = \left\lfloor \dfrac{1}{4} \right\rfloor \cdot 4 = 0 

 f_2 = \left\lfloor \dfrac{2}{4} \right\rfloor \cdot 4 = 0 

 f_3 = \left\lfloor \dfrac{3}{4} \right\rfloor \cdot 4 = 0 

 f_4 = \left\lfloor \dfrac{4}{4} \right\rfloor \cdot 4 = 4 

7. Let a_k = 2k + 1 and b_k = (k - 1)^3 + k + 2 for every integer k \geq 0. Show that the first three terms of these sequences are identical but that their fourth terms differ.

 a_0 = 2(0) + 1 = 1 

 a_1 = 2(1) + 1 = 3 

 a_2 = 2(2) + 1 = 5 

 a_3 = 2(3) + 1 = 7 

 b_0 = (0 - 1)^3 + 0 + 2 = 1 

 b_1 = (1 - 1)^3 + 1 + 2 = 3 

 b_2 = (2 - 1)^3 + 2 + 2 = 5 

 b_3 = (3 - 1)^3 + 3 + 2 = 13 

Compute the first fifteen terms of each of the sequences in 8 and 9, and describe the general behavior of these sequences in words. (A definition of logarithm is given in Section 7.1.)

8. g_n = \lfloor \log_{2}n \rfloor for every integer n \geq 1.

 g_1 = \lfloor \log_{2}(1) \rfloor = 0 

 g_2 = \lfloor \log_{2}(2) \rfloor = 1 

 g_3 = \lfloor \log_{2}(3) \rfloor = 1 

 g_4 = \lfloor \log_{2}(4) \rfloor = 2 

 g_5 = \lfloor \log_{2}(5) \rfloor = 2 

 g_6 = \lfloor \log_{2}(6) \rfloor = 2 

 g_7 = \lfloor \log_{2}(7) \rfloor = 2 

 g_8 = \lfloor \log_{2}(8) \rfloor = 3 

 g_9 = \lfloor \log_{2}(9) \rfloor = 3 

 g_{10} = \lfloor \log_{2}(10) \rfloor = 3 

 g_{11} = \lfloor \log_{2}(11) \rfloor = 3 

 g_{12} = \lfloor \log_{2}(12) \rfloor = 3 

 g_{13} = \lfloor \log_{2}(13) \rfloor = 3 

 g_{14} = \lfloor \log_{2}(14) \rfloor = 3 

 g_{15} = \lfloor \log_{2}(15) \rfloor = 3 

The general behavior of this sequence is that it increments in binary increments, as in it increments every 1, then 2, then 4, then 8 iterations of the index n.

9 h_n = n\lfloor \log_{2}n \rfloor for every integer n \geq 1.

 h_1 = (1)\lfloor \log_{2}(1) \rfloor = 0 

 h_2 = (2)\lfloor \log_{2}(2) \rfloor = 2 

 h_3 = (3)\lfloor \log_{2}(3) \rfloor = 3 

 h_4 = (4)\lfloor \log_{2}(4) \rfloor = 8 

 h_5 = (5)\lfloor \log_{2}(5) \rfloor = 10 

 h_6 = (6)\lfloor \log_{2}(6) \rfloor = 12 

 h_7 = (7)\lfloor \log_{2}(7) \rfloor = 14 

 h_8 = (8)\lfloor \log_{2}(8) \rfloor = 24 

 h_9 = (9)\lfloor \log_{2}(9) \rfloor = 27 

 h_{10} = (10)\lfloor \log_{2}(10) \rfloor = 30 

 h_{11} = (11)\lfloor \log_{2}(11) \rfloor = 33 

 h_{12} = (12)\lfloor \log_{2}(12) \rfloor = 36 

 h_{13} = (13)\lfloor \log_{2}(13) \rfloor = 39 

 h_{14} = (14)\lfloor \log_{2}(14) \rfloor = 42 

 h_{15} = (15)\lfloor \log_{2}(15) \rfloor = 45 

The sequence finds the minimal (floor) power of log_{2}n and then multiplies it by n, which is why there are sudden "jumps" when the floor calculates a jump to the next power of 2. For example, at n = 7 to n = 8, there is a noticeable jump because \lfloor \log_{2}7 \rfloor is 2, and then \lfloor \log_{2}8 \rfloor is 3.

Find explicit formulas for sequences of the form a_1, a_2, a_3, \dots with the initial terms given in 10-16.

10. -1, 1, -1, 1, -1, 1

a_n = (-1)^n where n is an integer such that n \geq 1.

11. 0, 1, -2, 3, -4, 5

a_n = (n - 1)(-1)^{n} where n is an integer such that n \geq 1.

12. \dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}

a_n = \dfrac{n}{(n + 1)^2} where n is an integer such that n \geq 1.

13. 1 - \dfrac{1}{2}, \dfrac{1}{2} - \dfrac{1}{3}, \dfrac{1}{3} - \dfrac{1}{4}, \dfrac{1}{4} - \dfrac{1}{5}, \dfrac{1}{5} - \dfrac{1}{6}, \dfrac{1}{6} - \dfrac{1}{7}

a_n = \dfrac{1}{n} - \dfrac{1}{n + 1} where n is an integer such that n \geq 1.

14. \dfrac{1}{3}, \dfrac{4}{9}, \dfrac{9}{27}, \dfrac{16}{81}, \dfrac{25}{243}, \dfrac{36}{729}

a_n = \dfrac{n^2}{3^n} where n is an integer such that n \geq 1.

15. 0, -\dfrac{1}{2}, \dfrac{2}{3}, -\dfrac{3}{4}, \dfrac{4}{5}, -\dfrac{5}{6}, \dfrac{6}{7}

a_n = \dfrac{(n - 1)(-1)^{n + 1}}{n} where n is an integer such that n \geq 1.

16. 3, 6, 12, 24, 48, 96

a_n = 3 \cdot 2^{n - 1} where n is an integer such that n \geq 1.

17. Consider the sequence defined by a_n = \dfrac{2n + (-1)^n - 1}{4} for every integer n \geq 0. Find an alternative explicit formula for a_n that uses the floor notation.

Omitted.

18. Let a_0 = 2, a_1 = 3, a_2 = -2, a_3 = 1, a_4 = 0, a_5 = -1, \text{ and } a_6 = -2. Compute each of the summations and products below.

a. \sum_{i = 0}^{6}{a_i}

 \sum_{i = 0}^{6}{a_i} = 2 + 3 + (-2) + 1 + 0 + (-1) + (-2) = 1 

b. \sum_{i = 0}^{0}{a_i}

 \sum_{i = 0}^{0}{a_i} = 2 

c. \sum_{j = 1}^{3}{a_{2j}}

 \sum_{j = 1}^{3}{a_{2j}} = (-2) + 0 + (-2) = -4 

d. \prod_{k = 0}^{6}{a_k}

 \prod_{k = 0}^{6}{a_k} = 2 \cdot 3 \cdot (-2) \cdot 1 \cdot 0 \cdot (-1) \cdot (-2) = 0 

e. \prod_{k = 2}^{2}{a_k}

 \prod_{k = 2}^{2}{a_k} = -2 

Compute the summations and products in 19-28.

19. \sum_{k = 1}^{5}{(k + 1)}

 \sum_{k = 1}^{5}{(k + 1)} = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) = 20 

20. \prod_{k = 2}^{4}{k^2}

 \prod_{k = 2}^{4}{k^2} = (2)^2 \cdot (3)^2 \cdot (4)^2 = 576 

21. \sum_{k = 1}^{3}{(k^2 + 1)}

 \sum_{k = 1}^{3}{(k^2 + 1)} = ((1)^2 + 1) + ((2)^2 + 1) + ((3)^2 + 1) = 17 

22. \prod_{j = 0}^{4}{(-1)^j}

 \prod_{j = 0}^{4}{(-1)^j} = (-1)^{(0)} \cdot (-1)^{(1)} \cdot (-1)^{(2)} \cdot (-1)^{(3)} \cdot (-1)^{(4)} = 1 

23. \sum_{i = 1}^{1}{i(i + 1)}

 \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) = 2 

24. \sum_{j = 0}^{0}{(j + 2) \cdot 2^j}

 \sum_{j = 0}^{0}{(j + 2) \cdot 2^j} = (0 + 2) \cdot 2^0 = 2 

25. \prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)}

 \prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)} =  \left(1 - \dfrac{1}{2}\right) = \frac{1}{2} 

26. \sum_{k = -1}^{1}{(k^2 + 3)}

 \sum_{k = -1}^{1}{(k^2 + 3)} = ((-1)^2 + 3) + ((0)^2 + 3) + ((1)^2 + 3) = 11 

27. \sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)}

 \sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)} = \left(\dfrac{1}{(1)} - \dfrac{1}{(1) + 1}\right) + \left(\dfrac{1}{(2)} - \dfrac{1}{(2) + 1}\right) + \left(\dfrac{1}{(3)} - \dfrac{1}{(3) + 1}\right) + \left(\dfrac{1}{(4)} - \dfrac{1}{(4) + 1}\right) + \left(\dfrac{1}{(5)} - \dfrac{1}{(5) + 1}\right) + \left(\dfrac{1}{(6)} - \dfrac{1}{(6) + 1}\right) = \frac{6}{7} 

28. \prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}}

 \prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}} = \dfrac{(2)((2) + 2)}{((2) - 1) \cdot ((2) + 1)} + \dfrac{(3)((3) + 2)}{((3) - 1) \cdot ((3) + 1)} + \dfrac{(4)((4) + 2)}{((4) - 1) \cdot ((4) + 1)} + \dfrac{(5)((5) + 2)}{((5) - 1) \cdot ((5) + 1)} = \frac{35}{3} 

Write the summations in 29-32 in expanded form.

29. \sum_{i = 1}^{n}{(-2)^i}

 \sum_{i = 1}^{n}{(-2)^i} = (-2)^1 + (-2)^2 + (-2)^3 + \dots + (-2)^{n} 

30. \sum_{j = 1}^{n}{j(j + 1)}

 \sum_{j = 1}^{n}{j(j + 1)} = ((1)((1) + 1)) + ((2)((2) + 1)) + ((3)((3) + 1)) + \dots + ((n)((n) + 1)) = 2 + 6 + 12 + \dots n(n + 1) 

31. \sum_{k = 0}^{n + 1}{\dfrac{1}{k!}}

 \sum_{k = 0}^{n + 1}{\dfrac{1}{k!}} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dots + \dfrac{1}{(n + 1)!} 

32. \sum_{i = 1}^{k + 1}{i(i!)}

 \sum_{i = 1}^{k + 1}{i(i!)} = 1(1!) + 2(2!) + 3(3!) + \dots + (k + 1)((k + 1)!) 

Evaluate the summations and products in 33-36 for the indicated values of the variable.

33. \dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dots + \dfrac{1}{n^2}; n = 1

 \frac{1}{1^2} = 1 

34. 1(1!) + 2(2!) + 3(3!) + \dots + m(m!); m = 2

 1(1!) + 2(2!) = 5 

35. \left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) \dots \left(\dfrac{k}{k + 1}\right); k = 3

 \left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) = \frac{1}{4} 

36. \left(\dfrac{1 \cdot 2}{3 \cdot 4}\right)\left(\dfrac{4 \cdot 5}{6 \cdot 7}\right)\left(\dfrac{6 \cdot 7}{8 \cdot 9}\right) \dots \left(\dfrac{m \cdot (m + 1)}{(m + 2) \cdot (m + 3)}\right); m = 1

 \left(\frac{1 \cdot 2}{3 \cdot 4}\right) = \frac{2}{12} = \frac{1}{6} 

Write each of 37-39 as a single summation.

37. \sum_{i = 1}^{k}{i^3 + (k + 1)^3}

 \sum_{i = 1}^{k}{i^3 + (k + 1)^3} = \sum_{i = 1}^{k + 1}{i^3} 

38. \sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}}

 \sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}} = \sum_{k = 1}^{m + 1}{\dfrac{k}{k + 1}}  

39. \sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}}

 \sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}} = \sum_{m = 0}^{n + 1}{(m + 1)2^m} 

Rewrite 40-42 by separating off the final term.

40. \sum_{i = 1}^{k + 1}{i(i!)}

 \sum_{i = 1}^{k + 1}{i(i!)} = \sum_{i = 1}^{k}{i(i!) + (k + 1)(k + 1)!} 

41. \sum_{k = 1}^{m + 1}{k^2}

 \sum_{k = 1}^{m + 1}{k^2} = \sum_{k = 1}^{m}{k^2 + (m + 1)^2} 

42. \sum_{m = 1}^{n + 1}{m(m + 1)}

 \sum_{m = 1}^{n + 1}{m(m + 1)} = \sum_{m = 1}^{n}{m(m + 1) + (n + 1)(n + 2)} 

Write each of 43-52 using summation or product notation.

43. 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2

 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 

 \sum_{k = 1}^{7}{(-1)^{k + 1}k^2} 

44. (1^3 - 1) - (2^3 - 1) + (3^3 - 1) - (4^3 - 1) + (5^3 - 1)

 \sum_{k = 1}^{5}{(-1)^{k + 1}(k^3 - 1)} 

45. (2^2 - 1) \cdot (3^2 - 1) \cdot (4^2 - 1)

 \prod_{k = 2}^{4}{(k^2 - 1)} 

46. \dfrac{2}{3 \cdot 4} - \dfrac{3}{4 \cdot 5} + \dfrac{4}{5 \cdot 6} - \dfrac{5}{6 \cdot 7} + \dfrac{6}{7 \cdot 8}

 \sum_{k=2}^{6}{\dfrac{(-1)^k \cdot k}{(k + 1) \cdot (k + 2)}} 

47. 1 - r + r^2 - r^3 + r^4 - r^5

 \sum_{k = 0}^{5}{(-1)^kr^{k + 1}} 

48. (1 - t) \cdot (1 - t^2) \cdot (1 - t^3) \cdot (1 - t^4)

 \prod_{k = 1}^{4}{(1 - t^k)} 

49. 1^3 + 2^3 + 3^3 + \dots + n^3

 \sum_{k}^{n}{k^3} 

50. \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dots + \dfrac{n}{(n + 1)!}

 \sum_{k = 1}^{n}{\frac{k}{(k + 1)!}} 

51. n + (n - 1) + (n - 2) + \dots + 1

 \sum_{k = 0}^{n - 1}{(n - k)} 

52. n + \dfrac{n - 1}{2!} + \dfrac{n - 2}{3!} + \dfrac{n - 3}{4!} + \dots + \dfrac{1}{n!}

 \sum_{k = 0}^{n - 1}{\frac{n - k}{(k + 1)!}} 

Transform each of 53 and 54 by making the change of variable i = k + 1.

 i = k + 1 

 i - 1 = k 

53. \sum_{k = 0}^{5}{k(k - 1)}

 \sum_{k = 0}^{5}{k(k - 1)} = \sum_{i = 1}^{6}{(i - 1)(i - 2)} 

54. \prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}}

 \prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}} = \prod_{i = 2}^{n + 1}{\frac{i - 1}{(i - 1)^2 + 4}} 

Transform each of 55-58 by making the change of variable j = i - 1.

 j = i - 1 

 i = j + 1 

55. \sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}}

 \sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}} = \sum_{j = 0}^{n}{\frac{j^2}{jn + n}} 

56. \sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}}

 \sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}} = \sum_{j = 2}^{n - 1}{\frac{j + 1}{j + n}} 

57. \sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}}

 \sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}} = \sum_{j = 0}^{n - 2}{\frac{j + 1}{(n - j - 1)^2}} 

58. \prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}}

 \prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}} = \prod_{j = n - 1}^{2n - 1}{\frac{n - j}{n + j + 1}} 

Write each of 59-61 as a single summation or product.

59. 3 \cdot \sum_{k = 1}^{n}{(2k - 3)} + \sum_{k = 1}^{n}{(4 - 5k)}

 \sum_{k = 1}^{n}{3(2k - 3) + (4 - 5k)} 

 \sum_{k = 1}^{n}{(6k - 9 + 4 - 5k)} 

 \sum_{k = 1}^{n}{(k - 5)} 

60. 2 \cdot \sum_{k = 1}^{n}{(3k^2 + 4)} + 5 \cdot \sum_{k = 1}^{n}{(2k^2 - 1)}

 \sum_{k = 1}^{n}{2(3k^2 + 4) + 5(2k^2 - 1)} 

 \sum_{k = 1}^{n}{(6k^2 + 8 + 10k^2 - 5)} 

 \sum_{k = 1}^{n}{(16k^2 + 3)} 

61. \left(\prod_{k = 1}^{n}{\dfrac{k}{k + 1}}\right) \cdot \left(\prod_{k = 1}^{n}{\dfrac{k + 1}{k + 2}}\right)

 \prod_{k = 1}^{n}{\left(\frac{k}{k + 1}\right)\left(\frac{k + 1}{k + 2}\right)} 

 \prod_{k = 1}^{n}{\left(\frac{k(k + 1)}{(k + 1)(k + 2)}\right)} 

 \prod_{k = 1}^{n}{\left(\frac{k\cancel{(k + 1)}}{\cancel{(k + 1)}(k + 2)}\right)} 

 \prod_{k = 1}^{n}{\left(\frac{k}{k + 2}\right)} 

Compute each of 62-76. Assume the values of the variables are restricted so that the expressions are defined.

62. \dfrac{4!}{3!}

 \frac{4!}{3!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 4 

63. \dfrac{6!}{8!}

 \frac{6!}{8!} = \frac{6!}{8 \cdot 7 \cdot 6!} = \frac{1}{56} 

64. \dfrac{4!}{0!}

 \frac{4!}{0!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{1} = 24 

65. \dfrac{n!}{(n - 1)!}

 \frac{n!}{(n - 1)!} = \frac{n(n - 1)!}{(n - 1)!} = n 

66. \dfrac{(n - 1)!}{(n + 1)!}

 \frac{(n - 1)!}{(n + 1)!} = \frac{(n - 1)!}{(n + 1)(n)(n - 1)!} = \frac{1}{n(n + 1)} 

67. \dfrac{n!}{(n - 2)!}

 \dfrac{n!}{(n - 2)!} = \frac{n(n - 1)(n - 2)!}{(n - 2)!} = n(n - 1) 

68. \dfrac{((n + 1)!)^2}{(n!)^2}

 \frac{((n + 1)!)^2}{(n!)^2} = \frac{((n + 1)(n!))^2}{(n!)^2} = (n + 1)^2 

69. \dfrac{n!}{(n - k)!}

 (n - k)! = (n - k)(n - k - 1)(n - k - 2) \dots (2)(1) 

 n! = n(n - 1)(n - 2) \dots (n - k + 1)(n - k)(n - k - 1)(n - k - 2) \dots (2)(1) 

 \frac{n!}{(n - k)!} = n(n - 1)(n - 2) \dots (n - k + 1) 

70. \dfrac{n!}{(n - k + 1)!}

 (n - k + 1)! = (n - k + 1)(n - k)(n - k - 1) \dots (2)(1) 

 n! = n(n - 1)(n - 2) \dots (n - k + 2)(n - k + 1)(n - k)(n - k - 1) \dots (2)(1)

 \frac{n!}{(n - k + 1)!} = n(n - 1)(n - 2) \dots (n - k + 2) 

71. \dbinom{5}{3}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{5}{3} = \frac{5!}{3!(5 - 3)!} 

 = \frac{5 \cdot 4 \cdot 3!}{3!(2)!} 

 = \frac{20}{2 \cdot 1} 

 = 10 

72. \dbinom{7}{4}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{7}{4} = \frac{7!}{4!(7 - 4)!} 

 = \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!(3)!} 

 = \frac{210}{3!} 

 = \frac{210}{6} 

 = 35 

73. \dbinom{3}{0}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{3}{0} = \frac{3!}{0!(3 - 0)!} 

 = \frac{3!}{1(3)!} 

 = 1 

74. \dbinom{5}{5}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{5}{5} = \frac{5!}{5!(5 - 5)!} 

 = \frac{1}{1(0)!} 

 = 1 

75. \dbinom{n}{n - 1}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{n}{n - 1} = \frac{n!}{(n - 1)!(n - (n - 1))!} 

 = \frac{n!}{(n - 1)!(n - n + 1)!} 

 = \frac{n!}{(n - 1)!(1)!} 

 = \frac{n(n - 1)!}{(n - 1)!(1)!} 

 = \frac{n}{1} 

 = n 

76. \dbinom{n + 1}{n - 1}

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

 \binom{n + 1}{n - 1} = \frac{(n + 1)!}{(n - 1)!((n + 1) - (n - 1))!} 

 = \frac{(n + 1)!}{(n - 1)!(n + 1 - n + 1)!} 

 = \frac{(n + 1)!}{(n - 1)!(2)!} 

 = \frac{(n + 1)(n)(n - 1)!}{(n - 1)!(2)!} 

 = \frac{n(n + 1)}{2} 

a. Prove that n! + 2 is divisible by 2, for every integer n \geq 2.

Proof:

Suppose that n is any integer such that n \geq 2.

By the definition of a factorial:

 n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 

Since n \geq 2, this can be represented as:

n! = \begin{cases} 2 & \text{if } n = 2 \ 3 \cdot 2 \cdot 1& \text{if } n = 3 \ n \cdot (n - 1) \dots \cdot 2 \cdot 1 & \text{if } n > 3 \ \end{cases}

In each case, n! has a factor of 2. Then:

 n! + 2 = 2k + 2 

 n! + 2 = 2(k + 1) 

for some integer k.

Now, k + 1 is an integer by the sum of integers.

Therefore n! + 2 is divisible by 2.

Q.E.D.

b. Prove that n! + k is divisible by k, for every integer n \geq 2 and k = 2, 3, \dots, n.

Proof:

Suppose n is any integer such that n \geq 2, and k is any integer such that 2 \leq k \leq n.

Since 2 \leq k \leq n, it follows that k is one of the factors of n!. Then:

 n! = km 

for some integer m.

By substitution:

 n! + k = km + k 

 = k(m + 1) 

Now, m + 1 is an integer by the sum of integers.

Therefore n! + k is divisible by k.

Q.E.D.

c. Given any integer m \geq 2, is it possible to find a sequence of m - 1 consecutive positive integers none of which is prime? Explain your answer.

Proof:

Suppose m is any integer such that m \geq 2.

Consider the sequence

 m! + 2, m! + 3, \dots, m! + m 

This is a sequence of m - 1 consecutive positive integers.

Let k be any integer such that 2 \leq k \leq m. The $k - 1$^th term of the sequence is m! + k.

Since k \leq m, it follows that k \mid m! (by part b). Then:

 m! = kt 

for some integer t.

Then:

 m! + k = kt + k = k(t + 1) 

Now, t + 1 is an integer by the sum of integers. Thus k divides m! + k and since k \geq 2 and (t + 1) > 1 are both factors greater than or equal to 1, it follows that m! + k is composite.

Therefore every term in the sequence is not prime, so there exists a sequence of m - 1 consecutive positive integers none of which is prime.

Q.E.D.

78. Prove that for all nonnegative integers n and r with

 r + 1 \leq n, \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} 

Proof:

Suppose n and r are any nonnegative integers such that r + 1 \leq n.

The given equation shown is:

 \frac{n - r}{r + 1}\binom{n}{r} = \frac{n - r}{r + 1}\left(\frac{n!}{r!(n - r)!}\right) 

 = \frac{n!(n - r)}{r!(r + 1)(n - r)!}

 = \frac{n!(n - r)}{r!(r + 1)(n - r)(n - r - 1)!}

 = \frac{n!}{r!(r + 1)(n - r - 1)!}

 = \frac{n!}{r!(r + 1)(n - (r + 1))!}

Notice that this in the form of a "n choose $r + 1$":

 \binom{n}{r + 1} 

Therefore, it has been shown that:

 \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} 

Q.E.D.

79. Prove that if p is a prime number and r is an integer with 0 < r < p, then \dbinom{p}{r} is divisible by p.

Proof:

Suppose that p is any prime number and r is any integer such that 0 < r < p.

[We need to show that p \mid \dbinom{p}{r}.]

Consider:

 \binom{p}{r} = \frac{p!}{r!(p - r)!} 

Since 0 < r < p, both r! and (p - r)! are less than p. Thus, the denominator r!(p - r)! can never have a factor of p.

The numerator can be expressed as p! = p(p - 1)!:

 \binom{p}{r} = \frac{p(p - 1)!}{r!(p - r)!} 

Factoring p out of the numerator gives:

 \binom{p}{r} = p \cdot \frac{(p - 1)!}{r!(p - r)!} 

Therefore it has been shown that:

 p \mid \binom{p}{r} 

Q.E.D.

80. Suppose a[1], a[2], a[3], \dots, a[m] is a one-dimensional array and consider the following algorithm segment:

\text{sum } := 0\\ \text{\textbf{for }} k := 1 \text{\textbf{ to }} m\\ \ \ \text{sum } := \text{ sum } + a[k]\\ \text{\textbf{next }} k

Fill in the blanks below so that each algorithm segment performs the same job as the one shown in the exercise statement.

a.

\text{sum } := 0\\ \text{\textbf{for }} i := 0 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} i

m - 1; \text{sum } + a[i + 1]

b.

\text{sum } := 0\\ \text{\textbf{for }} j := 2 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} j

m + 1; \text{sum } + a[j - 1]

Use repeated division by 2 to convert (by hand) the integers in 81-83 from base 10 to base 2.

81. 90

 90_{10} = 1011010_2 

82. 98

 98_{10} = 1100010_2 

83. 205

 205_{10} = 11001101_2 

Make a trace table to trace the action of Algorithm 5.1.1 on the input in 84-86.

84. 23

	0	1	2	3	4	5
a	23
r[i]		1	1	1	0	1
q	23	11	5	2	1	0
i	0	1	2	3	4	5

Outputs: 10111, which is 23_{10} = 10111_2.

85. 28

	0	1	2	3	4	5
a	28
r[i]		0	0	1	1	1
q	28	14	7	3	1	0
i	0	1	2	3	4	5

Outputs: 11100, which is 28_{10} = 11100_2.

86. 44

	0	1	2	3	4	5	6
a	44
r[i]		0	0	1	1	0	1
q	44	22	11	5	2	1	0
i	0	1	2	3	4	5	6

Outputs: 101100, which is 44_{10} = 101100_2

87. Write an informal description of an algorithm (using repeated division by 16) to convert a nonnegative integer from decimal notation to hexadecimal notation (base 16).

Input: a [a nonnegative integer]

Algorithm Body:

q := a, i := 0

[Repeatedly perform the integer division of q by 16 until q becomes 0. Store successive remainders in a one-dimensional array r[0], r[1], r[2], \dots r[k]. Even if the initial-value of q equals 0, the loop should execute one time (so that r[0] is computed). Thus the guard condition for the while loop is i = 0 or q \neq 0.]

\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 16\\ \ \ q := q \text{ div } 16\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}

[After execution of this step, the values of r[0], r[1], \dots, r[i - 1] are all 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, and a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_{16}.]

Output: r[0], r[1], r[2], \dots, r[i - 1] [a sequence of integers]

Use the algorithm you developed for exercise 87 to convert the integers in 88-90 to hexadecimal notation.

88. 287

 287_{10} = 11F_{16} 

89. 693

 693_{10} = 1BF_{16} 

91. 2,301

 2301_{10} = 8FD_{16} 

91. Write a formal version of the algorithm you developed for exercise 87.

Already done.

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Exercise Set 5.2

Page 309

1. Use the technique illustrated at the beginning of this section to show that the statements in (a) and (b) are true.

a. If \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5} then \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6}.

Since:

 \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5} 

then we can say that:

 \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6} = \frac{1}{5}\left(1 - \dfrac{1}{6}\right) 

Evaluating this right hand side, we find that:

 \frac{1}{5}\left(1 - \frac{1}{6}\right) 

 = \frac{1}{5}\left(\frac{6}{6} - \frac{1}{6}\right) 

 = \frac{1}{5}\left(\frac{5}{6}\right) 

 = \frac{1}{6} 

Which is equal to the right hand side of the equality to be proved.

b. If \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6} then \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \dfrac{1}{7}.

Given that:

 \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right) = \dfrac{1}{6} 

Then, by substitution:

 \left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right)\left(1 - \dfrac{1}{6}\right)\left(1 - \dfrac{1}{7}\right) = \frac{1}{6}\left(1 - \dfrac{1}{7}\right) 

Evaluating this right hand side, we find:

 \frac{1}{6}\left(1 - \frac{1}{7}\right) 

 = \frac{1}{6}\left(\frac{7}{7} - \frac{1}{7}\right) 

 = \frac{1}{6}\left(\frac{6}{7}\right) 

 = \frac{1}{7} 

And this is equal to the right hand side of the equality, and therefore shows that the statement is true.

2. For each positive integer n, let P(n) be the formula

 1 + 3 + 5 + \dots + (2n - 1) = n^2 

a. Write P(1). Is P(1) true?

 P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2  

By 5.2.1:

 P(n) = \frac{(2n - 1)((2n - 1) + 1)}{2} 

 = \frac{(2n - 1)(2n)}{2} 

 = \frac{4n^2 - 2n}{2} 

 = 2n^2 - n 

 P(1) = 1 + 3 + 5 + \dots + (2(1) - 1) = (1)^2  

 = 2(1)^2 - (1) = (1)^2  

 = 2(1) - (1) = (1)  

 = 2 - 1 = 1 

 = 1 = 1 

P(1) is true.

b. Write P(k).

 P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2  

 P(k) = 1 + 3 + 5 + \dots + (2k - 1) = k^2  

c. Write P(k + 1).

 P(n) = 1 + 3 + 5 + \dots + (2(n) - 1) = (n)^2  

 P(k + 1) = 1 + 3 + 5 + \dots + (2(k + 1) - 1) = (k + 1)^2 

Alternatively:

 P(k + 1) = 1 + 3 + 5 + \dots + (2k + 2 - 1) = k^2 + 2k + 1 

 P(k + 1) = 1 + 3 + 5 + \dots + (2k + 1) = k^2 + 2k + 1 

d. In a proof by mathematical induction that the formula holds for every integer n \geq 1, what must be shown in the inductive step?

In a proof by mathematical induction, where P(n) holds for every integer n \geq 1, the inductive step where for some integer k where it is assumed 1 + 3 + 5 + \dots + (2k - 1) = k^2 is true (inductive hypothesis), then 1 + 3 + 5 + \dots + (2(k + 1) - 1) = (k + 1)^2 must be shown to also be true.

3. For each positive integer n, let P(n) be the formula

 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} 

a. Write P(1). Is P(1) true?

 P(n) = 1^2 + 2^2 + \dots + (n)^2 = \frac{(n)((n) + 1)(2(n) + 1)}{6} 

By 5.2.1:

 P(n) = \frac{(n^2)((n^2) + 1)}{2} 

Then:

 P(1) = 1^2 + 2^2 + \dots + (1)^2 = \frac{(1)((1) + 1)(2(1) + 1)}{6} 

 = 1^2 + 2^2 + \dots + (1)^2 = \frac{(1)((1) + 1)(2(1) + 1)}{6} 

 = \frac{((1)^2)(((1)^2) + 1)}{2}= \frac{(1)((1) + 1)(2(1) + 1)}{6} 

 = \frac{(1)(1 + 1)}{2}= \frac{(1)(2)(2 + 1)}{6} 

 = \frac{(1)(2)}{2}= \frac{(1)(2)(3)}{6} 

 = \frac{2}{2} = \frac{6}{6} 

 = 1 = 1 

P(1) is true.

b. Write P(k).

 P(k) = 1^2 + 2^2 + \dots + k^2 = \frac{(k)(k + 1)(2k + 1)}{6} 

c. Write P(k + 1).

 P(k + 1) = 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} 

d. In a proof by mathematical induction that the formula holds for every integer n \geq 1, what must be shown in the inductive step?

In a proof by mathematical induction, where P(n) holds for every integer n \geq 1, the inductive step where for some integer k where it is assumed 1^2 + 2^2 + \dots + k^2 = \frac{(k)(k + 1)(2k + 1)}{6} is true (inductive hypothesis), then 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} must be shown to also be true.

4. For each integer n with n \geq 2, let P(n) be the formula

 \sum_{i = 1}^{n - 1}{i(i + 1)} = \frac{n(n - 1)(n + 1)}{3} 

a. Write P(2). Is P(2) true?

 P(n) = \sum_{i = 1}^{(n) - 1}{i(i + 1)} = \frac{(n)((n) - 1)((n) + 1)}{3} 

 P(2) = \sum_{i = 1}^{(2) - 1}{i(i + 1)} = \frac{(2)((2) - 1)((2) + 1)}{3} 

Compute left-hand side:

 \sum_{i = 1}^{(2) - 1}{i(i + 1)} 

 \sum_{i = 1}^{1}{i(i + 1)} 

 \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) 

 = (1)(2) 

 = 2 

Compute right-hand side:

 \frac{(2)((2) - 1)((2) + 1)}{3} 

 = \frac{(2)(1)(3)}{3} 

 = \frac{6}{3} 

 = 2 

Since both the left hand side and the right hand side are equal, P(2) is true.

b. Write P(k).

 P(k) = \sum_{i = 1}^{(k) - 1}{i(i + 1)} = \frac{(k)((k) - 1)((k) + 1)}{3} 

c. Write P(k + 1).

 P(k + 1) = \sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \frac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3} 

d. In a proof by mathematical induction that the formula holds for every integer n \geq 2, what must be shown in the inductive step?

In a proof by mathematical induction, where P(n) holds for every integer n \geq 2, the inductive step where for some integer k where it is assumed \sum_{i = 1}^{(k) - 1}{i(i + 1)} = \dfrac{(k)((k) - 1)((k) + 1)}{3} is true (inductive hypothesis), then \sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \dfrac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3} must be shown to also be true.

5. Fill in the missing pieces in the following proof that

 1 + 3 + 5 + \dots + (2n - 1) = n^2 

for every integer n \geq 1.

Proof: Let the property P(n) be the equation

 1 + 3 + 5 + \dots + (2n - 1) = n^2 

Show that P(1) is true:

To establish P(1), we must show that when 1 is substituted in place of n, the left-hand side equals the right-hand side. But when n = 1, the left-hand side is the sum of all the odd integers from 1 to 2 \cdot 1 - 1, which is the sum of the odd integers from 1 to 1 and is just 1. The right-hand side is __ (a) __, which also equals 1. So P(1) is true.

Show that for every integer k \geq 1, if P(k) is true then P(k + 1) is true:

Let k be any integer with k \geq 1.

[Suppose P(k) is true. That is:]

Suppose

1 + 3 + 5 \cdot + (2k - 1) = __ (b) __.

[This is the inductive hypothesis.]

[We must show that P(k + 1) is true. That is:]

We must show that __ c __ = __ (d) __.

Now the left-hand side of P(k + 1) is

 1 + 3 + 5 + \dots + (2(k + 1) - 1) 

 = 1 + 3 + 5 + \dots + (2k + 1) 

 = [1 + 3 + 5 + \dots + (2k - 1)] + (2k + 1) 

the next-to-last term is 2k - 1 because __ (e) __

 = k^2 + (2k + 1) 

by __ (f) __

 = (k + 1)^2 

which is the right-hand side of P(k + 1) [as was to be shown].

[Since we have proved the basis step and the inductive step, we conclude that the given statement is true.]

Note: This proof was annotated to help make its logical flow more obvious. In standard mathematical writing, such annotation is omitted.

a. (1)^2

b. k^2

c. 1 + 3 + 5 + \dots + (2(k + 1) - 1)

d. (k + 1)^2

e. the odd integer just before 2k + 1 is 2k - 1

f. inductive hypothesis

Prove each statement in 6-9 using mathematical induction. Do not derive them from Theorem 5.2.1 or Theorem 5.2.2.

6. For every integer n \geq 1,

 2 + 4 + 6 + \dots + 2n = n^2 + n 

Proof (by mathematical induction):

Let P(n) be the equation

 2 + 4 + 6 + \dots + 2n = n^2 + n 

Basis Step: Show that P(1) is true:

To establish P(1), we must show that when 1 is substituted in place of n, the left-hand side equals the right-hand side.

When n = 1, the left-hand side is the sum of all even integers from 2 to 2(1), which is the sum of the even integers from 2 to 2 and is just 2.

The right-hand side is 1^2 + 1, which also equals 2.

Therefore P(1) is true.

Inductive Step:

Show that for every integer k \geq 1, if P(k) is true then P(k + 1) is true:

Let k be any integer with k \geq 1.

Suppose P(k) is true. That is, suppose:

 2 + 4 + 6 + \dots + 2k = k^2 + k 

This is the inductive hypothesis.

We must show that P(k + 1) is true. That is we must show that:

 2 + 4 + 6 + \dots + 2(k + 1) = (k + 1)^2 + (k + 1) 

Now the left-hand side of P(k + 1) is

 2 + 4 + 6 + \dots + 2(k + 1) 

 = [2 + 4 + 6 + \dots + 2k] + (2(k + 1)) 

Where 2k is the next-to-last even term before 2k + 1. Then, by inductive hypothesis:

 = (k^2 + k) + (2(k + 1)) 

Then, by algebra:

 = k^2 + 3k + 2 

Now, the right-hand side is:

 (k + 1)^2 + (k + 1) 

 (k + 1)(k + 1) + (k + 1) 

 (k^2 + 2k + 1) + (k + 1) 

 k^2 + 3k + 2 

Thus, the left-hand and right-hand sides of P(k + 1) are equal. Hence P(k + 1) is true.

Since we have proved the basis step and the inductive step, we conclude that P(n) is true for every integer n \geq 1.

Q.E.D.

7. For every integer n \geq 1,

 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} 

Proof (by mathematical induction):

Let P(n) be the equation

 1 + 6 + 11 + 16 + \dots + (5n - 4) = \frac{n(5n - 3)}{2} 

Basis Step:

We must prove P(1):

 1 + 6 + 11 + 16 + \dots + (5(1) - 4) = \frac{(1)(5(1) - 3)}{2} 

When n = 1, the left-hand side is the sum of every fifth integer from 1 to 5(1) - 4, which is 1.

The right-hand side is:

 \frac{(1)(5(1) - 3)}{2} 

 = \frac{1(5 - 3)}{2} 

 = \frac{1(2)}{2} 

 = 1 

Both sides of the equality of P(1) are 1. So P(1) is true.

Inductive Step:

Let k be any integer with k \geq 1.

Suppose that P(k) is true. That is:

 1 + 6 + 11 + 16 + \dots + (5k - 4) = \frac{k(5k - 3)}{2} 

We must show that P(k + 1) is true. That is:

 1 + 6 + 11 + 16 + \dots + (5(k + 1) - 4) = \frac{(k + 1)(5(k + 1) - 3)}{2} 

Evaluating the left-hand side:

 1 + 6 + 11 + 16 + \dots + (5(k + 1) - 4) 

 = [1 + 6 + 11 + 16 + \dots + (5k - 4)] + (5(k + 1) - 4) 

Then, by inductive hypothesis:

 = \frac{k(5k - 3)}{2} + (5(k + 1) - 4) 

Then by algebra:

 = \frac{5k^2 - 3k}{2} + (5k + 5 - 4) 

 = \frac{5k^2 - 3k}{2} + \frac{2(5k + 5 - 4)}{2} 

 = \frac{5k^2 - 3k + 2(5k + 5 - 4)}{2} 

 = \frac{5k^2 - 3k + 10k + 10 - 8}{2} 

 = \frac{5k^2 + 7k + 2}{2} 

Now, the right-hand side:

 \frac{(k + 1)(5(k + 1) - 3)}{2} 

 = \frac{(k + 1)(5k + 5 - 3)}{2} 

 = \frac{5k^2 + 5k + 5k + 5 - 3k - 3}{2} 

 = \frac{5k^2 + 10k + 5 - 3k - 3}{2} 

 = \frac{5k^2 + 7k + 5 - 3}{2} 

 = \frac{5k^2 + 7k + 2}{2} 

which is the left-hand side of P(k + 1). Therefore P(k + 1) is true.

Q.E.D.

8. For every integer n \geq 0,

 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 

Proof (by mathematical induction):

Let P(n) be the equation:

 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 

Basis Step:

Prove P(0) is true.

 P(0) = 1 + 2 + 2^2 + \dots + 2^(0) = 2^{(0) + 1} - 1 

Evaluate the left-hand side when n = 0:

 1 + 2 + 2^2 + \dots + 2^(0) = 2^0 = 1 \quad \text{ when } n = 0 

Evaluate the right-hand side when n = 0:

 2^{(0) + 1} - 1 

 2^1 - 1 

 1 

Both the left-hand and right-hand sides of P(0) are equal. P(0) is true.

Inductive Step:

Let k be any integer with k \geq 0.

Suppose P(k) is true. That is:

 P(k) = 1 + 2 + 2^2 + \dots + 2^k = 2^{k + 1} + 1  

Prove that P(k + 1) is true:

 P(k + 1) = 1 + 2 + 2^2 + \dots + 2^(k + 1) = 2^{(k + 1) + 1} + 1  

 1 + 2 + 2^2 + \dots + 2^(k + 1) = 2^{(k + 1) + 1} + 1  

Evaluate the left-hand side:

 1 + 2 + 2^2 + \dots + 2^(k + 1) 

 [1 + 2 + 2^2 + \dots + 2^k] + 2^(k + 1) 

By inductive hypothesis:

 (2^{k + 1} + 1) + 2^(k + 1) 

 2(2^{k + 1}) + 1 

 2^{k + 2} + 1 

Evaluate the right-hand side:

 2^{(k + 1) + 1} + 1  

 = 2^{k + 2} + 1  

Therefore P(k + 1) is true.

Q.E.D.

9. For every integer n \geq 3,

 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} 

Proof by mathematical induction:

Let P(n) be the equation:

 4^3 + 4^4 + 4^5 + \dots + 4^n = \frac{4(4^n - 16)}{3} 

Basis Step:

Prove P(3). That is:

 4^3 + 4^4 + 4^5 + \dots + 4^3 = \frac{4(4^3 - 16)}{3} 

Evaluate left-hand side when n = 3:

 4^3 + 4^4 + 4^5 + \dots + 4^3 = 4^3 = 64 \quad \text{ when } n = 3 

Evaluate right-hand side when n = 3:

 \frac{4(4^3 - 16)}{3} 

 = \frac{4(64 - 16)}{3} 

 = \frac{4(48)}{3} 

 = \frac{192}{3} 

 = 64 

Therefore P(3) is true.

Inductive Step:

Let k be any integer where k \geq 3.

Suppose P(k). That is:

 4^3 + 4^4 + 4^5 + \dots + 4^k = \frac{4(4^k - 16)}{3} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 4^3 + 4^4 + 4^5 + \dots + 4^{k + 1} = \frac{4(4^{k + 1} - 16)}{3} 

Evaluate left-hand side:

 4^3 + 4^4 + 4^5 + \dots + 4^{k + 1} 

 = [4^3 + 4^4 + 4^5 + \dots + 4^k] + 4^{k + 1} 

By inductive hypothesis:

 = \frac{4(4^k - 16)}{3} + 4^{k + 1} 

 = \frac{4^{k + 1} - 64}{3} + \frac{3(4^{k + 1})}{3} 

 = \frac{4^{k + 1} - 64 + (3(4^{k + 1}))}{3} 

 = \frac{4^{k + 1} + 3(4^{k + 1}) - 64}{3} 

 = \frac{1(4^{k + 1}) + 3(4^{k + 1}) - 64}{3} 

 = \frac{4(4^{k + 1}) - 64}{3} 

 = \frac{4(4^{k + 1} - 16)}{3} 

Evaluate right-hand side:

 \frac{4(4^{k + 1} - 16)}{3} 

Both the left-hand and right-hand sides of P(k + 1) are equal. P(k + 1) is true.

Q.E.D.

Prove each of the statements in 10-18 by mathematical induction.

10. 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}, for every integer n \geq 1.

Proof (by mathematical induction):

Let P(n) be the equation:

 1^2 + 2^2 + \dots + n^2 = \dfrac{n(n + 1)(2n + 1)}{6} 

Basis Step:

Prove P(1). That is:

 1^2 + 2^2 + \dots + (1)^2 = \dfrac{(1)(1 + 1)(2(1) + 1)}{6} 

Evaluate left-hand side when n = 1:

 1^2 + 2^2 + \dots + (1)^2 = 1 

Evaluate right-hand side when n = 1:

 \dfrac{(1)(1 + 1)(2(1) + 1)}{6} 

 = \dfrac{(1)(2)(2 + 1)}{6} 

 = \dfrac{(1)(2)(3)}{6} 

 = \dfrac{6}{6} 

 = 1 

Both the left-hand and right-hand sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 1^2 + 2^2 + \dots + k^2 = \dfrac{k(k + 1)(2k + 1)}{6} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} 

 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)(k + 2)(2k + 2 + 1)}{6} 

 1^2 + 2^2 + \dots + (k + 1)^2 = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} 

Evaluate left-hand side:

 1^2 + 2^2 + \dots + (k + 1)^2 

 = [1^2 + 2^2 + \dots + k^2] + (k + 1)^2 

By inductive hypothesis:

 = \dfrac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 

 = \dfrac{k(k + 1)(2k + 1)}{6} + \frac{6(k + 1)^2}{6} 

 = \dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} 

 = \dfrac{(k + 1)[k(2k + 1) + 6(k + 1)]}{6} 

 = \dfrac{(k + 1)[2k^2 + k + 6k + 6]}{6} 

 = \dfrac{(k + 1)[2k^2 + 7k + 6]}{6} 

 = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} 

Evaluate right-hand side:

 = \dfrac{(k + 1)(k + 2)(2k + 3)}{6} 

Both the left-hand and right-hand sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

11. 1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2, for every integer n \geq 1.

Proof (by mathematical induction):

Let P(n) be the equation:

 1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n + 1)}{2}\right]^2 

Basis Step:

Prove P(1). That is:

 1^3 + 2^3 + \dots + (1)^3 = \left[\dfrac{(1)((1) + 1)}{2}\right]^2 

Evaluate left-hand when n = 1:

 1^3 + 2^3 + \dots + (1)^3 = 1 

Evaluate right-hand when n = 1:

 \left[\dfrac{(1)((1) + 1)}{2}\right]^2 

 = \left[\dfrac{(1)(2)}{2}\right]^2 

 = \left[\dfrac{2}{2}\right]^2 

 = [1]^2 

 = 1 

Both the left and right hand sides of P(1) are true. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 1^3 + 2^3 + \dots + k^3 = \left[\dfrac{k(k + 1)}{2}\right]^2 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 1^3 + 2^3 + \dots + (k + 1)^3 = \left[\dfrac{(k + 1)((k + 1) + 1)}{2}\right]^2 

Evaluate left-hand:

 1^3 + 2^3 + \dots + (k + 1)^3 

 = [1^3 + 2^3 + \dots + k^3] + (k + 1)^3 

By inductive hypothesis:

 = \left[\dfrac{k(k + 1)}{2}\right]^2 + (k + 1)^3 

 = \dfrac{k^2(k + 1)^2}{4} + (k + 1)^3 

 = \dfrac{k^2(k + 1)^2}{4} + \frac{4(k + 1)^3}{4} 

 = \dfrac{k^2(k + 1)^2 + 4(k + 1)^3}{4} 

 = \dfrac{k^2(k + 1)^2 + 4(k + 1)^2(k + 1)}{4} 

 = \dfrac{(k + 1)^2[k^2 + 4(k + 1)]}{4} 

 = \dfrac{(k + 1)^2[k^2 + 4k + 4]}{4} 

 = \dfrac{(k + 1)^2(k + 2)^2}{4} 

 = \left[\dfrac{(k + 1)(k + 2)}{2}\right]^2 

Evaluate right-hand:

 \left[\dfrac{(k + 1)((k + 1) + 1)}{2}\right]^2 

 = \left[\dfrac{(k + 1)(k + 2)}{2}\right]^2 

Both the left and right hand sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

12. \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1}, for every integer n \geq 1.

Proof (by mathematical induction):

Let P(n) be the equation:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1} 

Basis Step:

Prove P(1), that is:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(1)((1) + 1)} = \dfrac{(1)}{(1) + 1} 

Evaluate left-hand when n = 1:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(1)((1) + 1)} = \frac{1}{2} 

Evaluate right-hand when n = 1:

 \dfrac{(1)}{(1) + 1} 

 = \dfrac{1}{2} 

The left and right hand sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{k(k + 1)} = \dfrac{k}{k + 1} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)((k + 1) + 1)} = \dfrac{(k + 1)}{(k + 1) + 1} 

Alternatively:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)(k + 2)} = \dfrac{k + 1}{k + 2} 

Evaluate left-hand:

 \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \dfrac{1}{(k + 1)(k + 2)} 

 = \left[\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dots + \frac{1}{k(k + 1)}\right] + \dfrac{1}{(k + 1)(k + 2)} 

By the inductive hypothesis:

 = \dfrac{k}{k + 1} + \dfrac{1}{(k + 1)(k + 2)} 

 = \dfrac{k(k + 2)}{(k + 1)(k + 2)} + \dfrac{1}{(k + 1)(k + 2)} 

 = \dfrac{k(k + 2) + 1}{(k + 1)(k + 2)} 

 = \dfrac{k^2 + 2k + 1}{(k + 1)(k + 2)} 

 = \dfrac{(k + 1)(k + 1)}{(k + 1)(k + 2)} 

 = \dfrac{k + 1}{k + 2} 

Evaluate right-hand:

 \dfrac{k + 1}{k + 2} 

Both the left and right hand sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

13. \sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3}, for every integer n \geq 2.

Let P(n) be the equation:

 \sum_{i = 1}^{n - 1}{i(i + 1)} = \dfrac{n(n - 1)(n + 1)}{3} 

Basis Step:

Prove P(2). That is:

 \sum_{i = 1}^{(2) - 1}{i(i + 1)} = \dfrac{(2)((2) - 1)((2) + 1)}{3} 

Alternatively:

 \sum_{i = 1}^{1}{i(i + 1)} = \dfrac{(2)(1)(3)}{3} 

 \sum_{i = 1}^{1}{i(i + 1)} = 2 

Evaluate left-hand when n = 2:

 \sum_{i = 1}^{1}{i(i + 1)} 

 = (1)(1 + 1) = 2 

The left and right hand sides of P(2) are equal. Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 \sum_{i = 1}^{k - 1}{i(i + 1)} = \dfrac{k(k - 1)(k + 1)}{3} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \sum_{i = 1}^{(k + 1) - 1}{i(i + 1)} = \dfrac{(k + 1)((k + 1) - 1)((k + 1) + 1)}{3} 

Alternatively:

 \sum_{i = 1}^{k}{i(i + 1)} = \dfrac{(k + 1)(k)(k + 2)}{3} 

 \sum_{i = 1}^{k}{i(i + 1)} = \dfrac{k(k + 1)(k + 2)}{3} 

Evaluate left-hand:

 \sum_{i = 1}^{k}{i(i + 1)} 

 = \left[\sum_{i = 1}^{k - 1}{i(i + 1)}\right] + k(k + 1) 

By the inductive hypothesis:

 = \dfrac{k(k - 1)(k + 1)}{3} + k(k + 1) 

 = \dfrac{k(k - 1)(k + 1)}{3} + \frac{3k(k + 1)}{3} 

 = \dfrac{k(k - 1)(k + 1) + 3k(k + 1)}{3} 

 = \dfrac{k(k + 1)((k - 1) + 3)}{3} 

 = \dfrac{k(k + 1)(k + 2)}{3} 

Evaluate right-hand:

 \dfrac{k(k + 1)(k + 2)}{3} 

The left and right hand sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

14. \sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the equation:

 \sum_{i = 1}^{n + 1}{i \cdot 2^i} = n \cdot 2^{n + 2} + 2 

Basis Step:

Prove P(0). That is:

 \sum_{i = 1}^{(0) + 1}{i \cdot 2^i} = (0) \cdot 2^{(0) + 2} + 2 

Alternatively:

 \sum_{i = 1}^{1}{i \cdot 2^i} = (0) \cdot 2^{2} + 2 

 \sum_{i = 1}^{1}{i \cdot 2^i} = 0 + 2 

 \sum_{i = 1}^{1}{i \cdot 2^i} = 2 

Evaluate left-hand when n = 0:

 \sum_{i = 1}^{1}{i \cdot 2^i} 

 = (1) \cdot 2^(1) 

 = 2 

Both the left and right hand sides of P(0) are equal. Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 \sum_{i = 1}^{k + 1}{i \cdot 2^i} = k \cdot 2^{k + 2} + 2 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \sum_{i = 1}^{(k + 1) + 1}{i \cdot 2^i} = (k + 1) \cdot 2^{(k + 1) + 2} + 2 

Alternatively:

 \sum_{i = 1}^{k + 2}{i \cdot 2^i} = (k + 1) \cdot 2^{k + 3} + 2 

Evaluate left-hand:

 \sum_{i = 1}^{k + 2}{i \cdot 2^i} 

 = \left[\sum_{i = 1}^{k + 1}{i \cdot 2^i}\right] + (k + 2) \cdot 2^{k + 2} 

By the inductive hypothesis:

 = k \cdot 2^{k + 2} + 2 + (k + 2) \cdot 2^{k + 2} 

 = k(2^{k + 2}) + 2 + (k + 2)(2^{k + 2}) 

 = (2^{k + 2})(k + (k + 2)) + 2 

 = (2^{k + 2})(2k + 2) + 2 

 = 2(2^{k + 2})(k + 1) + 2 

 = (2^{k + 3})(k + 1) + 2 

 = (k + 1) \cdot 2^{k + 3} + 2 

Evaluate right-hand:

 (k + 1) \cdot 2^{k + 3} + 2 

The left and right hand sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

15. \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1, for every integer n \geq 1.

Proof (by mathematical induction):

Let P(n) be the equation:

 \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 

Basis Step:

Prove P(1). That is:

 \sum_{i = 1}^{(1)}{i(i!)} = ((1) + 1)! - 1 

Evaluate left-hand side:

 \sum_{i = 1}^{1}{i(i!)} 

 = 1(1!) = 1 

Evaluate right-hand side:

 ((1) + 1)! - 1 

 = (2)! - 1 

 = (2 \cdot 1) - 1 

 = 2 - 1 

 = 1 

Both sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 \sum_{i = 1}^{k}{i(i!)} = (k + 1)! - 1 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \sum_{i = 1}^{(k + 1)}{i(i!)} = ((k + 1) + 1)! - 1 

Alternatively:

 \sum_{i = 1}^{k + 1}{i(i!)} = (k + 2)! - 1 

Evaluate left-hand:

 \sum_{i = 1}^{k + 1}{i(i!)} 

 = \left[\sum_{i = 1}^{k}{i(i!)}\right] + (k + 1)(k + 1)! 

By the inductive hypothesis:

 = (k + 1)! - 1 + (k + 1)(k + 1)! 

 = (k + 1)! + (k + 1)(k + 1)! - 1 

 = (k + 1)!(1 + (k + 1)) - 1 

 = (k + 1)!(k + 2) - 1 

 = (k + 2)! - 1 

Evaluate right-hand:

 (k + 2)! - 1 

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

16. \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{n^2}\right) = \dfrac{n + 1}{2n}, for every integer n \geq 2.

Proof (by mathematical induction):

Let P(n) be the equation:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{n^2}\right) = \dfrac{n + 1}{2n} 

Basis Step:

Prove P(2). That is:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(2)^2}\right) = \dfrac{(2) + 1}{2(2)} 

Alternatively:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4} 

Evaluate left-hand side when n = 2:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{4}\right) 

 = \frac{3}{4} 

Both sides of P(2) are equal. Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{k^2}\right) = \dfrac{k + 1}{2k} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) = \dfrac{(k + 1) + 1}{2(k + 1)} 

Alternatively:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) = \dfrac{k + 2}{2k + 2} 

Evaluate left-hand:

 \left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \dfrac{1}{(k + 1)^2}\right) 

 = \left[\left(1 - \dfrac{1}{2^2}\right)\left(1 - \dfrac{1}{3^2}\right) \dots \left(1 - \frac{1}{k^2}\right)\right]\left(1 - \dfrac{1}{(k + 1)^2}\right) 

By the inductive hypothesis:

 = \left(\frac{k + 1}{2k}\right)\left(1 - \dfrac{1}{(k + 1)^2}\right) 

 = \left(\frac{k + 1}{2k}\right)\left(\frac{(k + 1)^2}{(k + 1)^2} - \dfrac{1}{(k + 1)^2}\right) 

 = \left(\frac{k + 1}{2k}\right)\left(\dfrac{(k + 1)^2 - 1}{(k + 1)^2}\right) 

 = \frac{(k + 1)((k + 1)^2 - 1)}{2k(k + 1)^2} 

 = \frac{(k + 1)^3 - (k + 1)}{2k(k + 1)^2} 

 = \frac{(k + 1)^2 - 1}{2k(k + 1)} 

 = \frac{(k + 1)(k + 1) - 1}{2k^2 + 2k} 

 = \frac{k^2 + 2k + 1 - 1}{2k^2 + 2k} 

 = \frac{k^2 + 2k}{2k^2 + 2k} 

 = \frac{k(k + 2)}{k(2k + 2)} 

 = \frac{k + 2}{2k + 2} 

Evaluate right-hand:

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

 \dfrac{k + 2}{2k + 2} 

17. \prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!}, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the equation:

 \prod_{i = 0}^{n}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2n + 2)!} 

Basis Step:

Prove P(0). That is:

 \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2(0) + 2)!} 

Alternatively:

 \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{2!} 

 \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{2} 

Evaluate left-hand when n = 0:

 \prod_{i = 0}^{(0)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} 

 = \frac{1}{2} 

Both sides of P(0) are equal. Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 \prod_{i = 0}^{k}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 2)!} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \prod_{i = 0}^{(k + 1)}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2(k + 1) + 2)!} 

Alternatively:

 \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 2 + 2)!} 

 \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} = \dfrac{1}{(2k + 4)!} 

Evaluate left-hand:

 \prod_{i = 0}^{k + 1}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)} 

 = \left[\prod_{i = 0}^{k}{\left(\dfrac{1}{2i + 1} \cdot \dfrac{1}{2i + 2}\right)}\right] \cdot \left(\frac{1}{2(k + 1) + 1} \cdot \frac{1}{2(k + 1) + 2}\right) 

By the inductive hypothesis:

 = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2(k + 1) + 1} \cdot \frac{1}{2(k + 1) + 2}\right) 

 = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2k + 2 + 1} \cdot \frac{1}{2k + 2 + 2}\right) 

 = \left(\frac{1}{(2k + 2)!}\right) \cdot \left(\frac{1}{2k + 3} \cdot \frac{1}{2k + 4}\right) 

 = \frac{1}{(2k + 4)!} 

Evaluate right-hand:

 \dfrac{1}{(2k + 4)!} 

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

18. \prod_{i = 2}^{n}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{n} for every integer n \geq 2.

Hint: See the discussion at the beginning of this section.

Proof (by mathematical induction):

Let P(n) be the equation:

 \prod_{i = 2}^{n}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{n} 

Basis Step:

Prove P(2). That is:

 \prod_{i = 2}^{2}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{2} 

Evaluate left-hand side when n = 2:

 \prod_{i = 2}^{2}{\left(1 - \dfrac{1}{i}\right)} 

 = 1 - \frac{1}{2} 

 = \frac{1}{2} 

Both sides of P(2) are equal. Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 \prod_{i = 2}^{k}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{k} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \prod_{i = 2}^{k + 1}{\left(1 - \dfrac{1}{i}\right)} = \dfrac{1}{k + 1} 

Evaluate left-hand side:

 \prod_{i = 2}^{k + 1}{\left(1 - \dfrac{1}{i}\right)} 

 = \left[\prod_{i = 2}^{k}{\left(1 - \dfrac{1}{i}\right)}\right] \cdot \left(1 - \frac{1}{k + 1}\right) 

By the inductive hypothesis:

 = \dfrac{1}{k} \cdot \left(1 - \frac{1}{k + 1}\right) 

 = \dfrac{1}{k} \cdot \left(\frac{k + 1}{k + 1} - \frac{1}{k + 1}\right) 

 = \dfrac{1}{k} \cdot \left(\frac{(k + 1) - 1}{k + 1}\right) 

 = \dfrac{1}{k} \cdot \left(\frac{k}{k + 1}\right) 

 = \frac{1}{k + 1} 

Evaluate right-hand side:

 \dfrac{1}{k + 1} 

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

19. (For students who have studied calculus) Use mathematical induction, the product rule from calculus, and the facts that \dfrac{d(x)}{dx} = 1 and that x^{k + 1} = x \cdot x^k to prove that for every integer n \geq 1, \dfrac{d(x^n)}{dx} = nx^{n - 1}.

Proof (by mathematical induction):

Let P(n) be the equation:

 \frac{d(x^n)}{dx} = nx^{n - 1} 

Basis Step:

Prove P(1). That is:

 \frac{d(x^{(1)})}{dx} = (1)x^{(1) - 1} 

Alternatively:

 \frac{dx}{dx} = 1x^0 

Evaluate the left-hand side when n = 1:

 \frac{dx}{dx} 

By the given fact that \dfrac{dx}{dx} = 1:

 = 1 

Evaluate the right-hand side when n = 1:

 = 1x^0 

 = 1 

Both the left and right hand sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 \frac{d(x^k)}{dx} = kx^{k - 1} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^{(k + 1) - 1} 

Alternatively:

 \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^k 

Evaluate left-hand side:

 \frac{d(x^{(k + 1)})}{dx} 

 \frac{d(x \cdot x^k)}{dx} 

By the product rule, we can separate this out into:

 \frac{dx}{dx} \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} 

By the given fact that \dfrac{dx}{dx} = 1:

 1 \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} 

By the inductive hypothesis:

 1 \cdot x^k + x \cdot kx^{k - 1}  

 x^k  + x \cdot kx^{k - 1}  

 x^k + kx^{k - 1 + 1}  

 x^k + kx^{k}  

 x^k(1 + k)  

 (k + 1)x^k  

Evaluate right-hand side:

 (k + 1)x^k 

Both the left and right sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

Use the formula for the sum of the first n integers and/or the formula for the sum of a geometric sequence to evaluate the sums in 20-29 or to write them in closed form.

20. 4 + 8 + 12 + 16 + \dots + 200

 4 + 8 + 12 + 16 + \dots + 200 

 = 4(1 + 2 + 3 + 4 + \dots + 50) 

 = 4\frac{50(51)}{2} 

 = 5100 

21. 5 + 10 + 15 + 20 + \dots + 300

 5 + 10 + 15 + 20 + \dots + 300 

 = 5(1 + 2 + 3 + 4 + \dots 60) 

 = 5\left(\frac{(60)(61)}{2}\right) 

 = 9150 

a. 3 + 4 + 5 + 6 + \dots + 1000

 3 + 4 + 5 + 6 + \dots + 1000 

 = (1 + 2 + 3 + 4 + \dots + 1000) - (1 + 2) 

 = \left(\frac{(1000)(1001)}{2}\right) - 3 

 = 500497 

b. 3 + 4 + 5 + 6 + \dots + m

 3 + 4 + 5 + 6 + \dots + m 

 = (1 + 2 + 3 + 4+ \dots + m) - (1 + 2) 

 = \left(\frac{(m)(m + 1)}{2}\right) - 3 

 = \frac{m^2 + m}{2} - 3 

 = \frac{m^2 + m}{2} - \frac{6}{2} 

 = \frac{m^2 + m - 6}{2} 

a. 7 + 8 + 9 + 10 + \dots + 600

 7 + 8 + 9 + 10 + \dots + 600 

 = (1 + 2 + 3 + 4 + \dots + 600) - (1 + 2 + 3 + 4 + 5 + 6) 

 = \left(\frac{(600)(601)}{2}\right) - 21 

 = 180279 

b. 7 + 8 + 9 + 10 + \dots + k

 7 + 8 + 9 + 10 + \dots + k 

 = (1 + 2 + 3 + 4 + \dots + k) - (1 + 2 + 3 + 4 + 5 + 6) 

 = \left(\frac{(k)(k + 1)}{2}\right) - 21 

 = \frac{k^2 + k}{2} - 21 

 = \frac{k^2 + k - 42}{2} 

24. 1 + 2 + 3 + \dots + (k - 1), where k is any integer with k \geq 2.

 1 + 2 + 3 + \dots + (k - 1) 

 = \frac{(k - 1)((k - 1) + 1)}{2} 

 = \frac{(k - 1)(k)}{2} 

 = \frac{k^2 - k}{2} 

a. 1 + 2 + 2^2 + \dots + 2^{25}

 1 + 2 + 2^2 + \dots + 2^{25} 

 = \frac{2^{25 + 1} - 1}{2^{25} - 1} 

 = \frac{2^{26} - 1}{2 - 1} 

 = 67108863 

b. 2 + 2^2 + 2^3 + \dots + 2^{26}

 2 + 2^2 + 2^3 + \dots + 2^{26} 

 k 2(1 + 2 + 2^2 + \dots + 2^{25}) 

By part a:

 = 2(67108863) 

 = 134217726 

c. 2 + 2^2 + 2^3 + \dots + 2^n

 2 + 2^2 + 2^3 + \dots + 2^n 

 2(1 + 2 + 2^2 + \dots + 2^{n - 1}) 

 2\left(\frac{2^{(n - 1) + 1} - 1}{2 - 1}\right) 

 2\left(\frac{2^n - 1}{1}\right) 

 2(2^n - 1) 

 2^{n + 1} - 2 

26. 3 + 3^2 + 3^3 + \dots + 3^n, where n is any integer with n \geq 1.

 3 + 3^2 + 3^3 + \dots + 3^n 

 3(1 + 3 + 3^2 + \dots + 3^{n - 1}) 

 3\left(\frac{3^{(n - 1) + 1} - 1}{3 - 1}\right) 

 3\left(\frac{3^n - 1}{2}\right) 

 \frac{3^{n + 1} - 3}{2} 

27. 5^3 + 5^4 + 5^5 + \dots + 5^k, where k is any integer with k \geq 3.

 5^3 + 5^4 + 5^5 + \dots + 5^k 

 = 5^3(1 + 5 + 5^2 + \dots + 5^{k - 3}) 

 = 5^3\left(\frac{5^{(k - 3) + 1} - 1}{5 - 1}\right) 

 = 5^3\left(\frac{5^{k - 2} - 1}{4}\right) 

 = \frac{5^{k - 2 + 3} - 5^3}{4} 

 = \frac{5^k - 5^3}{4} 

28. 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n}, where n is any positive integer.

 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n} 

 = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{\dfrac{1}{2} - 1} 

 = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{-\dfrac{1}{2}} 

 = \left[\left(\dfrac{1}{2}\right)^{n + 1} - 1\right](-2) 

 = \left(\dfrac{-2}{2}\right)^{n + 1} + 2 

 = 2 - \left(\dfrac{-2}{2}\right)^{n + 1} 

 = 2 + \dfrac{1}{2^n} 

29. 1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n, where n is any positive integer.

 1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n 

 = 1 + (-2) + (-2)^2 + (-2)^3 + \dots + (-2)^n 

 = \frac{(-2)^{n + 1} - 1}{(-2) - 1} 

 = \frac{(-2)^{n + 1} - 1}{-3} 

30. Observe that

 \frac{1}{1 \cdot 3} = \frac{1}{3} 

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} = \frac{2}{5} 

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = \frac{3}{7} 

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} = \frac{4}{9} 

Guess a general formula and prove it by mathematical induction.

General formula:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} 

for all integers n \geq 1.

Proof (by mathematical induction):

Let P(n) be the equation:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} 

Basis Step:

Prove P(1):

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} = \frac{(1)}{2(1) + 1} 

Evaluate left-hand side when n = 1:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} 

 = \frac{1}{(2 - 1)(2 + 1)}

 = \frac{1}{(1)(3)}

 = \frac{1}{3} 

Evaluate right-hand side when n = 1:

 \frac{(1)}{2(1) + 1} 

 \frac{1}{2 + 1} 

 \frac{1}{3} 

The left and right hand sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{(k + 1)}{2(k + 1) + 1} 

Alternatively:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 2 - 1)(2k + 2 + 1)} = \frac{k + 1}{2k + 2 + 1} 

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3} 

Evaluate the left-hand side:

 \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} 

 = \left[\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)}\right] + \frac{1}{(2k + 1)(2k + 3)} 

By the inductive hypothesis:

 = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} 

 = \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} 

 = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} 

 = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} 

 = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} 

 = \frac{k + 1}{2k + 3} 

Evaluate the right-hand side:

 \frac{k + 1}{2k + 3} 

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

31. Compute values of the product

 \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \dots \left(1 + \frac{1}{n}\right) 

for small values of n in order to conjecture a general formula for the product. Prove your conjecture by mathematical induction.

32. Observe that

 1 = 1 

 1 - 4 = -(1 + 2) 

 1 - 4 + 9 = 1 + 2 + 3 

 1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4) 

 1 - 4 + 9 - 16 + 25 = 1 + 2 + 3 + 4 + 5 

Guess a general formula and prove it by mathematical induction.

 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) 

Proof (by mathematical induction):

Let P(n) be the equation:

 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) 

for all integers n \geq 1.

Basis Step:

Prove P(1). That is:

 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) = (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) 

Evaluate left-hand side when n = 1:

 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) 

 = (-1)^{0}(1^2) 

 = 1(1) 

 = 1 

Evaluate right-hand side when n = 1:

 (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) 

 = 1 

Both sides of P(1) are equal. Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k). That is:

 1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2) = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 1 - 4 + 9 - 16 + \dots + (-1)^{(k + 1) - 1}((k + 1)^2) = (-1)^{(k + 1) - 1}(1 + 2 + 3 + 4 + \dots + (k + 1)) 

Alternatively:

 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) 

Evaluate left-hand:

 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) 

 = \left[1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2)\right] + (-1)^{k}((k + 1)^2) 

By the inductive hypothesis:

 = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) + (-1)^{k}((k + 1)^2) 

 = (-1)^{k - 1}\left[(1 + 2 + 3 + 4 + \dots + k) - (k + 1)^2\right] 

By 5.2.1:

 = (-1)^{k - 1}\left[\frac{k(k + 1)}{2} - (k + 1)^2\right] 

 = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - (k + 1)\right)\right] 

 = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - \frac{2(k + 1)}{2}\right)\right] 

 = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2(k + 1)}{2}\right)\right] 

 = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2k - 2)}{2}\right)\right] 

 = (-1)^{k - 1}\left[(k + 1)\left(\frac{-k - 2)}{2}\right)\right] 

 = (-1)^{k - 1}\left[(-1)(k + 1)\left(\frac{k + 2}{2}\right)\right] 

 = (-1)^{k - 1}\left[(-1)\left(\frac{(k + 1)(k + 2)}{2}\right)\right] 

 = (-1)^{k}\left(\frac{(k + 1)(k + 2)}{2}\right) 

By 5.2.1:

 = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) 

Evaluate right-hand:

 (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) 

Both sides of P(k + 1) are equal. Therefore P(k + 1) is true.

Q.E.D.

33. Find a formula in n, a, m, and d for the sum (a + md) + (a + (m + 1)d) + (a + (m + 2)d) + \dots + (a + (m + n)d), where m and n are integers, n \geq 0, and a and d are real numbers. Justify your answer.

 a + (a + d) + (a + 2d) + \dots (a + nd) = (n + 1)a + d\left(\frac{n(n + 1)}{2}\right) 

34. Find a formula in a, r, m, and n for the sum

 ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} 

where m and n are integers, n \geq 0, and a and r are real numbers. Justify your answer.

 ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} = ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) 

By factoring out the ar^m, this just becomes a geometric series:

 ar^m(1 + r + r^2 + r^3 + \dots r^n) 

And by 5.2.2, we can substitute that series out with:

 ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) 

35. You have two parents, four grandparents, eight great-grandparents, and so forth.

a. If all your ancestors were distinct, what would be the total number of your ancestors for the past 40 generations (counting your parents' generation as number one)? (Hint: Use the formula for the sum of a geometric sequence.)

The geometric sequence for this is:

 1 + 2 + 2^2 + 2^3 + \dots + 2^n 

So, by 5.2.2, this is:

 \frac{2^{n + 1} - 1}{2 - 1} 

Where n is the number of generations. Plugging in 39 (since we count as the first generation) returns:

 \frac{2^{39 + 1} - 1}{2 - 1} 

 = \frac{2^{40} - 1}{1} 

 = 2^{40} - 1 

 = 1099511627775 

b. Assuming that each generation represents 25 years, how long is 40 generations?

 25 \cdot 1099511627775 

 \approx 2.748779069 \cdot 10^{13} \text{ years} 

c. The total number of people who have ever lived is approximately 10 billion, which equals 10^{10} people. Compare this fact with the answer to part (a). What can you deduce?

When demarcated for easier reading, part a's answer reads as:

 = 1,099,511,627,775 

Which is 1 trillion, 99 billion, 511 million, 627 thousand, 775 people. Since this exceeds the approximate total number of people who have ever lived. We can deduce that some(probably many) of my ancestors must have been related to one another.

Find the mistakes in the proof fragments in 36-38.

Theorem:

For any integer n \geq 1,

 1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} 

"Proof (by mathematical induction):

Certainly the theorem is true for n = 1 because 1^2 = 1 and \dfrac{1(1 + 1)(2 \cdot 1 + 1)}{6} = 1 . So the basis step is true. For the inductive step, suppose that k is any integer with k \geq 1, k^2 = \dfrac{k(k + 1)(2k + 1)}{6}. We must show that (k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}."

In the inductive step, the inductive hypothesis reads:

 k^2 = \frac{k(k + 1)(2k + 1)}{6} 

But it should read:

 1^2 + 2^2 + \dots + k^2 = \frac{k(k + 1)(2k + 1)}{6} 

This error cascades into their proof, which reads:

 (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} 

But instead should read:

 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} 

Theorem:

For any integer n \geq 0,

 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 

"Proof (by mathematical induction):

Let the property P(n) be

 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 

Show that P(0) is true:

The left-hand side of P(0) is 1 + 2 + 2^2 + \dots + 2^0 = 1 and the right-hand side is 2^{0 + 1} - 1 = 2 - 1 = 1 also. So P(0) is true."

The left-hand side evaluation should instead read:

The left-hand side of P(0) is 2^0 = 1 since when n = 0, only the first term is evaluated..

Theorem:

For any integer n \geq 1,

 \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 

"Proof (by mathematical induction):

Let the property P(n) be

 \sum_{i = 1}^{n}{i(i!)} = (n + 1)! - 1 

Show that P(1) is true:

When n = 1,

 \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 

So

 1(1!) = 2! - 1

and

 1 = 1 

Thus P(1) is true."

The author of this proof fragment incorrectly rewrites the upper limit as i instead of 1. They write:

 \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 

When it should be:

 \sum_{i = 1}^{1}{i(i!)} = (1 + 1)! - 1 

Then, they should evaluate each side independently, but instead they simply evaluate each together, which is incorrect. Instead the basis step should be written as:

Evaluate the left-hand side when n = 1:

 \sum_{i = 1}^{1}{i(i!)} 

 = 1(1!) 

 = 1(1) 

 = 1 

Evaluate the right-hand side when n = 1:

 (1 + 1)! - 1 

 = (2)! - 1 

 = (2 \cdot 1) - 1 

 = 2 - 1 

 = 1 

Both sides of P(1) are equal. Therefore P(1) is true.

39. Use Theorem 5.2.1 to prove that if m and n are any positive integers and m is odd, then \sum_{k = 0}^{m - 1}{(n + k)} is divisible by m. Does the conclusion hold if m is even? Justify your answer.

Omitted.

40. Use Theorem 5.2.1 and the result of exercise 10 to prove that if p is any prime number with p \geq 5, then the sum of the squares of any p consecutive integers is divisible by p.

Omitted.

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Exercise Set 5.3

Page 320

1. Use mathematical induction (and the proof of proposition 5.3.1 as a model) to show that any amount of money of at least 14¢ can be made up using 3¢ and 8¢ coins.

Proof (by mathematical induction):

Let P(n) be the sentence:

$n$¢ can be obtained using $3$¢ and $8$¢ coins.

Basis Step:

Prove P(14):

P(14) is true because $14$¢ can be obtained using one $8$¢ coin and two $3$¢ coins.

Inductive Step:

Let k be any integer where k \geq 14.

Suppose P(k) is true. That is:

$k$¢ can be obtained using $3$¢ and $8$¢ coins.

Prove P(k + 1). That is:

$k + 1$¢ can be obtained using $3$¢ and $8$¢ coins.

Case 1 (there is a $8$¢ coin among those used to make up $k$¢):

In this case, replace the $8$¢ coin with three $3$¢ coins. The result will be $k + 1$¢.

Case 2 (there is not a $8$¢ coin among those used to make up $k$¢):

In this case, because k \geq 14, at least 5 $3$¢ coins must have been used. So remove five $3$¢ coins and replace them with two $8$¢ coins. The result will be $k + 1$¢.

Therefore in either case $(k + 1)$¢ can be obtained using $3$¢ and $8$¢ coins.

Q.E.D.

2. Use mathematical induction to show that any postage of at least 12¢ can be obtained using 3¢ and 7¢ stamps.

Proof (by mathematical induction):

Let P(n) be the sentence:

$n$¢ postage can be obtained using $3$¢ and $7$¢ stamps.

Basis Step:

Prove P(12). That is:

$12$¢ postage can be obtained using $3$¢ and $7$¢ stamps.

$12$¢ can be obtained using four $3$¢ stamps. Therefore P(12) is true.

Inductive Step:

Let k be any integer where k \geq 12.

Suppose P(k). That is:

$k$¢ postage can be obtained using $3$¢ and $7$¢ stamps.

Prove P(k + 1). That is:

$(k + 1)$¢ postage can be obtained using $3$¢ and $7$¢ stamps.

Case 1 (at least one 2 $3$¢ stamps are used to make up $k$¢):

Replace the two $3$¢ stamps with a $7$¢ stamp. This results in $(k + 1)$¢.

Case 2 (there are 1 or 0 $3$¢ stamps among those used to make up $k$¢):

Replace two $7$¢ stamps with five $3$¢ stamps. This results in $(k + 1)$¢.

Therefore, in both cases (k + 1) postage can be obtained using $3$¢ and $7$¢ stamps.

Q.E.D.

3. Stamps are sold in packages containing either 5 stamps or 8 stamps.

a. Show that a person can obtain 5, 8, 10, 13, 15, 16, 20, 21, 24, or 25 stamps by buying a collection of 5-stamp packages and 8-stamp packages.

• 5 stamps can be obtained by purchasing one 5 stamp package.

• 8 stamps can be obtained by purchasing one 8 stamp package.

• 10 stamps can be obtained by purchasing two 5 stamp packages.

• 13 stamps can be obtained by purchasing one 5 stamp package and one 8 stamp package.

• 15 stamps can be obtained by purchasing three 5 stamp packages.

• 16 stamps can be obtained by purchasing two 8 stamp packages.

• 20 stamps can be obtained by purchasing four 5 stamp packages.

• 21 stamps can be obtained by purchasing two 8 stamp packages and one 5 stamp package.

• 24 stamps can be obtained by purchasing three 8 stamp packages.

• 25 stamps can be obtained by purchasing five 5 stamp packages.

b. Use mathematical induction to show that any quantity of at least 28 stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

Proof (by mathematical induction):

Let P(n) be the sentence:

n stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

Basis Step:

Prove P(28). That is:

28 stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

28 stamps can be obtained by buying four 5-stamp packages and one 8-stamp package.

Therefore P(28) is true.

Inductive Step:

Let k be any integer where k \geq 28.

Suppose P(k). That is:

k stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

Prove P(k + 1). That is:

(k + 1) stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

Case 1 (at least three 5-stamp packages are used in obtaining k stamps):

Replace three 5-stamp packages with two 8-stamp packages. This results in (k + 1) stamps.

Case 2 (at most two 5-stamp packages are used in obtaining k stamps):

If there at most two 5-stamp packages, that means that 28-10=18 must be made up of 8-stamp packages. So at least 3 8-stamp packages must be used to exceed the 28 minimum.

Replace three 8-stamp packages with 5 5-stamp packages. This results in (k + 1) stamps.

Therefore in both cases (k + 1) stamps can be obtained by buying a collection of 5-stamp packages and 8-stamp packages.

Q.E.D.

4. For each positive integer n, let P(n) be the sentence that describes the following divisibility property:

 5^n - 1 \text{ is divisible by } 4 

a. Write P(0). Is P(0) true?

 5^0 - 1 = 1 - 1 = 0 

P(0) is true, as 0 = 0 \cdot 4.

b. Write P(k).

 P(k) = 5^k - 1 \text{ is divisible by } 4 

c. Write P(k + 1).

 P(k + 1) = 5^{k + 1} - 1 \text{ is divisible by } 4 

d. In a proof by mathematical induction that this divisibility property holds for every integer n \geq 0, what must be shown in the inductive step?

It must be shown that supposing that 5^k - 1 is divisible by 4 for some integer k \geq 0, that therefore 5^{k + 1} - 1 is divisible by 4.

5. For each positive integer n, let P(n) be the inequality

 2^n < (n + 1)! 

a. Write P(2). Is P(2) true?

 P(2) = 2^2 < (2 + 1)! 

 P(2) = 4 < (3)! 

 P(2) = 4 < (3 \cdot 2 \cdot 1) 

 P(2) = 4 < 6 

Yes, P(2) is true because 4 is less than 6.

b. Write P(k).

 P(k) = 2^k < (k + 1)! 

c. Write P(k + 1).

 P(k + 1) = 2^{k + 1} < ((k + 1) + 1)! 

Alternatively:

 P(k + 1) = 2^{k + 1} < (k + 2)! 

d. In a proof by mathematical induction that this inequality holds for every integer n \geq 2, what must be shown in the inductive step?

It must be shown that supposing 2^k < (k + 1)! is true for any integer k \geq 2, that therefore 2^{k + 1} < (k + 2)! is true.

6. For each positive integer n, let P(n) be the sentence

Any checkerboard with dimensions 2 \times 3n can be completely covered with L-shaped trominoes.

a. Write P(1). Is P(1) true?

Any checkerboard with dimensions 2 \times 3(1) can be completely covered with L-shaped trominoes.

Yes, this is true, a 2 \times 3 dimension checkerboard can be completely covered with L-shaped trominoes (2 in fact.)

b. Write P(k).

Any checkerboard with dimensions 2 \times 3k can be completely covered with L-shaped trominoes.

c. Write P(k + 1).

Any checkerboard with dimensions 2 \times 3(k + 1) can be completely covered with L-shaped trominoes.

d. In a proof by mathematical induction that P(n) is true for each integer n \geq 1, what must be shown in the inductive step?

It must be shown that supposing any checkerboard with dimensions 2 \times 3k can be completely covered with L-shaped trominoes for any integer k \geq 1, that therefore any checkerboard with dimensions 2 \times 3(k + 1) can be completely covered with L-shaped trominoes.

7. For each positive integer n, let P(n) be the sentence

In any round-robin tournament involving n teams, the teams can be labeled T_1, T_2, T_3, \dots, T_n, so that T_i beats T_{i + 1} for every i = 1, 2, \dots, n.

a. Write P(2). Is P(2) true?

In any round-robin tournament involving 2 teams, the teams can be labeled T_1, T_2, so that T_i beats T_{i + 1} for every i = 1, 2.

This is true, in a round-robin tournament involving only 2 teams, one can label the teams such that T_2 beats T_1.

b. Write P(k).

In any round-robin tournament involving k teams, the teams can be labeled T_1, T_2, T_3, \dots, T_k, so that T_i beats T_{i + 1} for every i = 1, 2, \dots, k.

c. Write P(k + 1).

In any round-robin tournament involving (k + 1) teams, the teams can be labeled T_1, T_2, T_3, \dots, T_{k + 1}, so that T_i beats T_{i + 1} for every i = 1, 2, \dots, (k + 1).

d. In a proof by mathematical induction that P(n) is true for each integer n \geq 2, what must be shown in the inductive step?

It must be shown that supposing in any round-robin tournament involving k teams, the teams can be labeled T_1, T_2, T_3, \dots T_k, so that T_i beats T_{i + 1} for every i = 1, 2, \dots k for any integer k \geq 2, then therefore in any round-robin tournament involving (k + 1) teams, the teams can be labeled T_1, T_2, T_3, \dots T_{k + 1} so that T_i beats T_{i + 1} for every i = 1, 2, \dots (k + 1).

Prove each statement in 8-23 by mathematical induction.

8. 5^n - 1 is divisible by 4, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 5^n - 1 \text{ is divisible by } 4 

Basis Step:

Prove P(0). That is:

 5^0 - 1 \text{ is divisible by } 4 

 1 - 1 \text{ is divisible by } 4 

 0 \text{ is divisible by } 4 

This sentence is true as 0 = 0 \cdot 4, which shows that 0 is divisible by 4 by the definition of divisibility.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 5^k - 1 \text{ is divisible by } 4 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 5^{k + 1} - 1 \text{ is divisible by } 4 

 5^{k + 1} - 1 

 = 5^k \cdot 5 - 1 

 = 5^k \cdot (4 + 1) - 1 

 = 5^k \cdot 4 + 5^k - 1 

Since we know by the inductive hypothesis that 5^k - 1 is divisible by 4. By the definition of divisibility:

 5^k - 1 = 4r 

for some integer r. Our equation now becomes:

 = 5^k \cdot 4 + 4r 

 = 4(5^k + r) 

Now, we know that 5^k + r is an integer by the sum and product of integers. Therefore, by the definition of divisibility, 5^{k + 1} - 1 is divisible by 4.

Q.E.D.

9. 7^n - 1 is divisible by 6, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 7^n - 1 \text{ is divisible by } 6 

Basis Step:

Prove P(0). That is:

 7^0 - 1 \text{ is divisible by } 6 

 7^0 - 1 

 = 1 - 1 

 = 0 

0 is divisible by 6 because 0 = 0 \cdot 6.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 7^k - 1 \text{ is divisible by } 6 

Prove P(k + 1). That is:

 7^{k + 1} - 1 \text{ is divisible by } 6 

 7^{k + 1} - 1 

 = 7^k \cdot 7 - 1 

 = 7^k \cdot (6 + 1) - 1 

 = 7^k \cdot 6 + (7^k - 1) 

By the inductive hypothesis and by the definition of divisibility:

 = 7^k \cdot 6 + 6r 

for some integer r.

 = 6(7^k + r) 

Now, we know that 7^k + r is an integer by the sum and product of integers. Therefore, by the definition of divisibility, 7^{k + 1} - 1 is divisible by 6.

Q.E.D.

10. n^3 - 7n + 3 is divisible by 3, for each integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 n^3 - 7n + 3 \text{ is divisible by } 3 

Basis Step:

Prove P(0). That is:

 (0)^3 - 7(0) + 3 \text{ is divisible by } 3 

 (0)^3 - 7(0) + 3 

 = 0 - 0 + 3 

 = 3 

By the definition of divisibility, 3 \mid 3, as 3 = 1 \cdot 3.

Therefore, P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 k^3 - 7k + 3 \text{ is divisible by } 3 

Prove P(k + 1). That is:

 (k + 1)^3 - 7(k + 1) + 3 \text{ is divisible by } 3 

 (k + 1)^3 - 7(k + 1) + 3 

 = (k + 1)(k + 1)(k + 1) - 7k - 7 + 3 

 = (k^2 + 2k + 1)(k + 1) - 7k - 7 + 3 

 = (k^2(k + 1) + 2k(k + 1) + 1(k + 1)) - 7k - 7 + 3 

 = (k^3 + k^2 + 2k^2 + 2k + k + 1) - 7k - 7 + 3 

 = (k^3 + 3k^2 + 3k + 1) - 7k - 7 + 3 

 = (k^3 - 7k + 3) + 3k^2 + 3k + 1 - 7 

 = (k^3 - 7k + 3) + 3k^2 + 3k - 6 

By the inductive hypothesis and definition of divisibility:

 = (3r) + 3k^2 + 3k - 6 

for some integer r.

 = 3(r + k^2 + k - 2) 

Now, we know that r + k^2 + k - 2 is an integer by the product and sum of integers. Thus, by the definition of divisibility, (k + 1)^3 - 7(k + 1) + 3 is divisible by 3.

Therefore P(k + 1) is true.

Q.E.D.

11. 3^{2n} - 1 is divisible by 8, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 3^{2n} - 1 \text{ is divisible by } 8 

Basis Step:

Prove P(0). That is:

 3^{2(0)} - 1 \text{ is divisible by } 8 

 3^{2(0)} - 1 

 = 3^0 - 1 

 = 1 - 1 

 = 0 

0 is divisible by 8 as 0 = 0 \cdot 8.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 3^{2k} - 1 \text{ is divisible by } 8 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 3^{2(k + 1)} - 1 \text{ is divisible by } 8 

 3^{2(k + 1)} - 1 

 = 3^{2k + 2} - 1 

 = 3^{2k} \cdot 3^2 - 1 

 = 3^{2k} \cdot 9 - 1 

 = 3^{2k} \cdot (8 + 1) - 1 

 = 3^{2k} \cdot 8 + (3^{2k} - 1) 

By the inductive hypothesis and the definition of divisibility:

 = 3^{2k} \cdot 8 + 8r 

for some integer r.

 = 8(3^{2k} + r) 

Now, 3^{2k} + r is an integer by the sum and product of integers. Thus 3^{2(k + 1)} - 1 is divisible by 8 by the definition of divisibility.

Therefore P(k + 1) is true.

Q.E.D.

12. For any integer n \geq 0, 7^n - 2^n is divisible by 5.

Proof (by mathematical induction):

Let P(n) be the sentence:

 7^n - 2^n \text{ is divisible by } 5 

Basis Step:

Prove P(0). That is:

 7^0 - 2^0 \text{ is divisible by } 5 

 7^0 - 2^0 

 = 1 - 1 

 = 0 

0 is divisible by 5 as 0 = 0 \cdot 5.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 7^k - 2^k \text{ is divisible by } 5 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 7^{k + 1} - 2^{k + 1} \text{ is divisible by } 5 

 7^{k + 1} - 2^{k + 1} 

 = 7^k \cdot 7^1 - 2^k \cdot 2^1 

 = 7^k \cdot (5 + 2) - 2^k \cdot 2^1 

 = 7^k \cdot 5 + (2)7^k - 2^k \cdot 2^1 

 = 7^k \cdot 5 + 2(7^k - 2^k) 

By the inductive hypothesis and the definition of divisibility:

 = 7^k \cdot 5 + 2(5r) 

For some integer r.

 = 5(7^k + 2r) 

Now, 7^k + 2r is an integer by the sum and product of integers. Thus 7^{k + 1} - 2^{k + 1} is divisible by 5 by the definition of divisibility.

Therefore P(k + 1) is true.

Q.E.D.

13. For any integer n \geq 0, x^n -y^n is divisible by x - y, where x and y are any integers with x \neq y.

Proof (by mathematical induction):

Suppose x and y are any integers with x \neq y.

Let P(n) be the sentence:

 x^n - y^n \text{ is divisible by } x - y 

Basis Step:

Prove P(0). That is:

 x^0 - y^0 \text{ is divisible by } x - y 

 x^0 - y^0 

 = 1 - 1 

 = 0 

0 is divisible by (x - y) as 0 = 0 \cdot (x - y).

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 x^k - y^k \text{ is divisible by } x - y 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 x^{k + 1} - y^{k + 1} \text{ is divisible by } x - y 

 x^{k + 1} - y^{k + 1} 

 = x^k(x) - y^k(y) 

 = x^k(x) - xy^k + xy^k - y^k(y) 

 = x(x^k - y^k) + y^k(x - y) 

By the inductive hypothesis:

 = x(r(x - y)) + y^k(x - y) 

for some integer r.

 = (x - y)(xr + y^k) 

We know xr + y^k is an integer by the sum and product of integers. By the definition of divisibility, x^{k + 1} - y^{k + 1} is divisible by x - y.

Therefore P(k + 1) is true.

Q.E.D.

14. n^3 - n is divisible by 6, for each integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 n^3 - n \text{ is divisible by } 6 

Basis Step:

Prove P(0). That is:

 0^3 - 0 \text{ is divisible by } 6 

 0^3 - 0 

 = 0 - 0 

 = 0 

0 is divisible by 6 because 0 = 0 \cdot 6.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 k^3 - k \text{ is divisible by } 6 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 (k + 1)^3 - (k + 1) \text{ is divisible by } 6 

 (k + 1)^3 - (k + 1) 

 = (k + 1)(k + 1)(k + 1) - (k + 1) 

 = (k^2 + 2k + 1)(k + 1) - (k + 1) 

 = (k^3 + k^2 + 2k^2 + 2k + k + 1) - (k + 1) 

 = (k^3 + 3k^2 + 3k + 1) - (k + 1) 

 = k^3 + 3k^2 + 3k + 1 - k - 1 

 = k^3 + 3k^2 + 2k 

 = (k^3 - k) + 3k^2 + 3k 

 = (k^3 - k) + 3k(k + 1) 

By the inductive hypothesis and definition of divisibility:

 = 6r + 3k(k + 1) 

for some integer r.

By Theorem 4.5.2, the product of any two consecutive integers must be even.

 = 6r + 3(2m) 

for some integer m.

 = 6r + 6m 

 = 6(r + m) 

Now, r + m is an integer by the sum of integers.

Therefore (k + 1)^3 - (k + 1) is divisible by 6 by the definition of divisibility.

Therefore P(k + 1) is true.

Q.E.D.

15. n(n^2 + 5) is divisible by 6, for each integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the sentence:

 n(n^2 + 5) \text{ is divisible by } 6 

Basis Step:

Prove P(0). That is:

 0(0^2 + 5) \text{ is divisible by } 6 

 0(0^2 + 5) 

 = 0(0 + 5) 

 = 0(5) 

 = 0 

0 is divisible by 6 as 0 = 0 \cdot 6.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 k(k^2 + 5) \text{ is divisible by } 6 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 (k + 1)((k + 1)^2 + 5) \text{ is divisible by } 6 

 (k + 1)((k + 1)^2 + 5) 

 = (k + 1)((k + 1)(k + 1) + 5) 

 = (k + 1)(k^2 + 2k + 6) 

 = k^3 + k^2 + 2k^2 + 2k + 6k + 6 

 = k^3 + 3k^2 + 8k + 6 

 = k^3 + 3k^2 + 5k + 3k + 6 

 = (k^3 + 5k) + 3k^2 + 3k + 6 

 = k(k^2 + 5) + 3k^2 + 3k + 6 

 = k(k^2 + 5) + 3(k^2 + k + 2) 

By the inductive hypothesis and definition of divisibility:

 = 6r + 3(k^2 + k + 2) 

for some integer r.

 = 6r + 3(k(k + 1) + 2) 

By Theorem 4.5.2 k(k + 1) is always even:

 = 6r + 3(2m + 2) 

for some integer m.

 = 6r + 6m + 6 

 = 6(r + m + 1) 

Now, r + m + 1 is an integer by the sum of integers. Thus (k + 1)((k + 1)^2 + 5) is divisible by 6 by the definition of divisibility.

Therefore P(k + 1) is true.

Q.E.D.

16. 2^n < (n + 1)!, for every integer n \geq 2.

Proof (by mathematical induction):

Let P(n) be the sentence:

 2^n < (n + 1)! 

Basis Step:

Prove P(2). That is:

 2^(2) < (2 + 1)! 

 4 < (3)! 

 4 < (3 \cdot 2 \cdot 1) 

 4 < 6 

4 is less than 6.

Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 2^k < (k + 1)! 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 2^{k + 1} < ((k + 1) + 1)! 

Alternatively:

 2^{k + 1} < (k + 2)! 

By the inductive hypothesis and the laws of exponents:

 = 2^{k} \cdot 2 < 2(k + 1)! 

Since k \geq 2, then 2 < k + 2, and so:

 2(k + 1)! < (k + 2)(k + 1)! = (k + 2)! 

Combining these inequalities shows:

 2^{k + 1} < (k + 2)! 

As was to be shown.

Q.E.D.

17. 1 + 3n \leq 4^n, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the inequality:

 1 + 3n \leq 4^n 

Basis Step:

Prove P(0). That is:

 1 + 3(0) \leq 4^0 

 = 1 + 0 \leq 1 

 = 1 \leq 1 

Since 1 = 1, 1 \leq 1 is a true statement.

Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 1 + 3k \leq 4^k 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 1 + 3(k + 1) \leq 4^{k + 1} 

 (1 + 3k) + 3 \leq 4^k \cdot 4 

By the inductive hypothesis:

 (1 + 3k) + 3 \leq 4^k + 3 

Now show:

 4^k + 3 \leq 4^{k + 1} 

Since:

 4^{k + 1} = 4^k \cdot 4 

it is enough to show:

 3 \leq 3 \cdot 4^k 

which is true for all k \geq 0.

So:

 1 + 3(k + 1) \leq 4^k + 3 \leq 4^{k + 1} 

 1 + 3(k + 1) \leq 4^{k + 1} 

Q.E.D.

18. 5^n + 9 < 6^n, for each integer n \geq 2.

Proof (by mathematical induction):

Let P(n) be the inequality:

 5^n + 9 < 6^n 

Basis Step:

Prove P(2). That is:

 5^2 + 9 < 6^2 

 25 + 9 < 36 

 34 < 36 

Since 34 is less than 36, this inequality is true.

Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 5^k + 9 < 6^k 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 5^{k + 1} + 9 < 6^{k + 1} 

If we multiply the inductive hypothesis by 5:

 5(5^k + 9) < 5(6^k) 

 5^{k + 1} + 45 < 5(6^k) 

 5^{k + 1} + 45 < 5(6^k) < 6^{k + 1} 

 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} 

Note that:

 5^{k + 1} + 9 < 5^{k + 1} + 45 < 5(6^k) < 6^k \cdot 6 = 6^{k + 1} 

Therefore:

 5^{k + 1} + 9 < 6^{k + 1} 

As was to be shown.

Q.E.D.

19. n^2 < 2^n, for every integer n \geq 5.

Proof (by mathematical induction):

Let P(n) be the inequality:

 n^2 < 2^n 

Basis Step:

Prove P(5). That is:

 5^2 < 2^5 

 25 < 32 

Since 25 is less than 32, this is a true statement.

Therefore P(5) is true.

Inductive Step:

Let k be any integer where k \geq 5.

Suppose P(k). That is:

 k^2 < 2^k 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 (k + 1)^2 < 2^{k + 1} 

Now, expanding out the left-hand side:

 (k + 1)^2 = k^2 + 2k + 1 

Consider the inductive hypothesis:

 k^2 < 2^k 

It follows that:

 k^2 + 2k + 1 < 2^k + 2k + 1 

By proposition 5.3.2, 2k + 1 < 2^k since k \geq 5 \geq 3.

Hence:

 (k + 1)^2 = k^2 + 2k + 1 < 2^k + 2k + 1 < 2^k + 2^k = 2^{k + 1} 

 (k + 1)^2 < 2^{k + 1} 

As was to be shown.

Q.E.D.

20. 2^n < (n + 2)!, for each integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the inequality:

 2^n < (n + 2)! 

Basis Step:

Prove P(0). That is:

 2^0 < (0 + 2)! 

 1 < (2)! 

 1 < 2 

Since 1 is less than 2. This is a true statement.

Therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 0.

Suppose P(k). That is:

 2^k < (k + 2)! 

Prove P(k + 1). That is:

 2^{k + 1} < ((k + 1) + 2)! 

Alternatively:

 2^{k + 1} < (k + 3)! 

Expanding out the left-hand side:

 2^{k + 1} = 2^k \cdot 2 

Consider the inductive hypothesis:

 2^k < (k + 2)! 

Multiple both sides by 2:

 2(2^k) < 2(k + 2)! 

 2^{k + 1} < 2(k + 2)! 

Now, expanding out the right-hand side:

 (k + 3)! = (k + 3)(k + 2)! 

Since k \geq 0, it follows that k + 3 \geq 3 \geq 2. Putting out inequalities together then, we get:

 2^{k + 1} < 2(k + 2)! < (k + 3)(k + 2)! = (k + 3)! 

And now simplified:

 2^{k + 1} < (k + 3)! 

As was to be shown.

Q.E.D.

21. \sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}}, for every integer n \geq 2.

Proof (by mathematical induction):

Let P(n) be the inequality:

 \sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{n}} 

Basis Step:

Prove P(2). That is:

 \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{2}} 

\dots \dfrac{1}{\sqrt{2}} just ends at term, \dfrac{1}{\sqrt{2}}.

 \sqrt{2} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} 

 \sqrt{2} < 1 + \dfrac{1}{\sqrt{2}} 

 \sqrt{2} < \frac{\sqrt{2}}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} 

 \sqrt{2} < \dfrac{\sqrt{2} + 1}{\sqrt{2}} 

 (\sqrt{2})(\sqrt{2}) < \left(\dfrac{\sqrt{2} + 1}{\sqrt{2}}\right)(\sqrt{2}) 

 2 < \sqrt{2} + 1 \approx 2.414213562 

This statement is true. Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} 

From the inductive hypothesis:

 \sqrt{k} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} 

Add \dfrac{1}{\sqrt{k + 1}} to both sides:

 \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} 

From here, it is enough to show:

 \sqrt{k + 1} \leq \sqrt{k} + \frac{1}{\sqrt{k + 1}} 

 \sqrt{k + 1} - \sqrt{k} \leq \frac{1}{\sqrt{k + 1}} 

 \left(\sqrt{k + 1} - \sqrt{k}\right)\left(\frac{\sqrt{k + 1} + \sqrt{k}}{\sqrt{k + 1} + \sqrt{k}}\right) \leq \frac{1}{\sqrt{k + 1}} 

 \frac{(k + 1) - k}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} 

 \frac{1}{\sqrt{k + 1} + \sqrt{k}} \leq \frac{1}{\sqrt{k + 1}} 

Since \sqrt{k + 1} + \sqrt{k} > \sqrt{k + 1}, this inequality holds.

Simplified, our inequality becomes:

 \sqrt{k + 1} < \sqrt{k} + \frac{1}{\sqrt{k + 1}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} 

 \sqrt{k + 1} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \dots + \dfrac{1}{\sqrt{k + 1}} 

As was to be shown.

Q.E.D.

22. 1 + nx \leq (1 + x)^n, for every real number x > -1 and every integer n \geq 2.

Proof (by mathematical induction):

Suppose x is any real number where x > -1.

Let P(n) be the sentence:

 1 + nx \leq (1 + x)^n 

Basis Step:

Prove P(2). That is:

 1 + 2x \leq (1 + x)^2 

 1 + 2x \leq 1 + 2x + x^2 

 0 \leq x^2 

This inequality always holds.

Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 1 + kx \leq (1 + x)^k 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 1 + (k + 1)x \leq (1 + x)^{k + 1} 

Consider the inductive hypothesis:

 1 + kx \leq (1 + x)^k 

Multiply each side by (1 + x):

 (1 + x)(1 + kx) \leq ((1 + x)^k)(1 + x) 

 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} 

Now it is enough to show that the left hand side of P(k + 1) is less than or equal to the left-hand side of (1 + x)(P(k)):

 1 + (k + 1)x \leq 1 + x + kx + kx^2 

 1 + kx + x  \leq 1 + x + kx + kx^2 

 1 + x + kx  \leq 1 + x + kx + kx^2 

 0 \leq kx^2 

Since k \geq 2, this inequality will always hold.

Simplified, our inequality is:

 1 + (k + 1)x \leq 1 + x + kx + kx^2 \leq (1 + x)^{k + 1} 

 1 + (k + 1)x \leq (1 + x)^{k + 1} 

As was to be shown.

Q.E.D.

a. n^3 > 2n + 1, for each integer n \geq 2.

Proof (by mathematical induction):

Let P(n) be the inequality:

 n^3 > 2n + 1 

Basis Step:

Prove P(2). That is:

 (2)^3 > 2(2) + 1 

 8 > 4 + 1 

 8 > 5 

Since 8 is greater than 5, this statement is true.

Therefore P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2.

Suppose P(k). That is:

 k^3 > 2k + 1 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 (k + 1)^3 > 2(k + 1) + 1 

Alternatively:

 (k + 1)^3 > 2k + 2 + 1 

 (k + 1)^3 > 2k + 3 

Consider the inductive hypothesis:

 k^3 > 2k + 1 

Add 2 to both sides:

 k^3 + 2 > 2k + 1 + 2 

 k^3 + 2 > 2k + 3 

Now it is enough to prove that the left-hand side of this inequality is less than the left-hand side of the P(k + 1) inequality:

 (k + 1)^3 > k^3 + 2 

 (k + 1)(k + 1)(k + 1) > k^3 + 2 

 (k^2 + 2k + 1)(k + 1) > k^3 + 2 

 k^3 + k^2 + 2k^2 + 2k + k + 1 > k^3 + 2 

 k^3 + 3k^2 + 3k + 1 > k^3 + 2 

 3k^2 + 3k > 1 

Since k \geq 2, this inequality will always hold.

Simplified:

 (k + 1)^3 > k^3 + 2 > 2k + 3 

 (k + 1)^3 > 2k + 3 

As was to be shown.

Q.E.D.

b. n! > n^2, for each integer n \geq 4.

Proof (by mathematical induction):

Let P(n) be the inequality:

 n! > n^2 

Basis Step:

Prove P(4). That is:

 4! > 4^2 

 (4 \cdot 3 \cdot 2 \cdot 1) > 16 

 24 > 16 

Since 24 is greater than 16, this statement is true.

Therefore P(4) is true.

Inductive Step:

Let k be any integer where k \geq 4.

Suppose P(k). That is:

 k! > k^2 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 (k + 1)! > (k + 1)^2 

Take the inductive hypothesis:

 k! > k^2 

And multiply each side by (k + 1):

 (k + 1)k! > k^2(k + 1) 

 (k + 1)! > k^2(k + 1) 

Now it is enough to show:

 k^2(k + 1) > (k + 1)^2 

 k^2 > k + 1 

And this inequality holds for all k \geq 4.

Simplified:

 (k + 1)! > k^2(k + 1) > (k + 1)^2 

 (k + 1)! > (k + 1)^2 

As was to be shown.

Q.E.D.

24. A sequence a_1, a_2, a_3, \dots is defined by letting a_1 = 3 and a_k = 7a_{k - 1} for each integer k \geq 2. Show that a_n = 3 \cdot 7^{n - 1} for every integer n \geq 1.

Proof (by mathematical induction):

Let P(n) be the statement:

 a_n = 3 \cdot 7^{n - 1} 

Basis Step:

Prove P(1). That is:

 a_1 = 3 \cdot 7^{1 - 1} 

 = 3 \cdot 7^{0} 

 = 3 \cdot 1 

 = 3 

Since a_1 = 3 is defined in the problem statement, this equality is true.

Therefore P(1) is true.

_Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k). That is:

 a_k = 3 \cdot 7^{k - 1} 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 a_{k + 1} = 3 \cdot 7^{(k + 1) - 1} 

Alternatively:

 a_{k + 1} = 3 \cdot 7^k 

By the definition of the given sequence:

 a_{k + 1} = 7a_k 

By the inductive hypothesis:

 = 7(3 \cdot 7^{k - 1}) 

By the laws of exponents:

 = 3 \cdot 7^k 

And this is the right-hand side of the equality to be shown.

Q.E.D.

25. A sequence b_0, b_1, b_2, \dots is defined by letting b_0 = 5 and b_k = 4 + b_{k - 1} for each integer k \geq 1. Show that b_n > 4n for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the inequality:

 b_n > 4n 

Basis Step:

Prove P(0). That is:

 b_0 > 4(0) 

 5 > 4(0) 

 5 > 0 

This inequality holds. Therefore P(0) is true.

Inductive Step:

Let k be any integer where k \geq 0.

Suppose P(k). That is:

 b_k > 4k 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 b_{k + 1} > 4(k + 1) 

By the definition of the sequence:

 b_{k + 1} = 4 + b_k 

By the inductive hypothesis:

 > 4 + 4k 

 > 4(1 + k) 

 > 4(k + 1) 

Q.E.D.

26. A sequence c_0, c_1, c_2, \dots is defined by letting c_0 = 3 and c_k = (c_{k - 1})^2 for every integer k \geq 1. Show that c_n = 3^{2n} for each integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the equality:

 c_n = 3^{2n} 

Basis Step:

Prove P(0). That is:

 c_0 = 3^{2(0)} 

 c_0 = 3^{0} 

 c_0 = 1 

Stopping here. It is likely Epp has a typo, she means c_n = 3^{2^n}, not c_n = 3^{2n}.

27. A sequence d_1, d_2, d_3, \dots is defined by letting d_1 = 2 and d_k = \dfrac{d_{k - 1}}{k} for each integer k \geq 2. Show that for every integer n \geq 1, d_n = \dfrac{2}{n!}.

Proof (by mathematical induction):

Let P(n) be the equation:

 d_n = \frac{2}{n!} 

Basis Step:

Prove P(1). That is:

 d_1 = \frac{2}{1!} 

 d_1 = \frac{2}{1} 

 d_1 = 2 

Since the problem statement states that d_1 = 2, this matches our equality.

Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 d_k = \frac{2}{k!} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 d_{k + 1} = \frac{2}{(k + 1)!} 

By the given sequence:

 d_{k + 1} = \frac{d_k}{k + 1} 

By the inductive hypothesis:

 = \frac{2}{(k + 1)k!} 

 = \frac{2}{(k + 1)!} 

Q.E.D.

28. Prove that for every integer n \geq 1,

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} 

Proof (by mathematical induction):

Let P(n) be the equality:

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2n - 1)}{(2n + 1) + (2n + 3) + \dots + (2n + (2n - 1))} 

Basis Step:

Prove P(1), that is:

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(1) - 1)}{(2(1) + 1) + (2(1) + 3) + \dots + (2(1) + (2(1) - 1))} 

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2 - 1)}{(2 + 1) + (2 + 3) + \dots + (2 + (2 - 1))} 

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + 1}{(2 + 1) + (2 + 3) + \dots + (2 + 1)} 

 \frac{1}{3} = \frac{1}{(2 + 1) + (2 + 3) + \dots + 3} 

 \frac{1}{3} = \frac{1}{3} 

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k), that is:

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2(k + 1) - 1)}{(2(k + 1) + 1) + (2(k + 1) + 3) + \dots + (2(k + 1) + (2(k + 1) - 1))} 

Starting from the inductive hypothesis:

 \frac{1}{3} = \frac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (2k + (2k - 1))} 

Omitted.

Exercises 29 and 30 use the definition of string and string length from page 13 in Section 1.4. Recursive definitions for these terms are given in section 5.9.

29. A set L consists of strings obtained by juxtaposing one or more of abb, bab, and bba. Use mathematical induction to prove that for every integer n \geq 1, if a string s in L has a length 3n, then s contains an even number of b's.

Proof (by mathematical induction):

Suppose a set L consists of strings by juxtaposing one or more of abb, bab, and bba.

Let P(n) be the sentence:

If a string s in L has length 3n, then s contains an even number of b's.

Basis Step:

Prove P(1), that is:

If a string s in L has length 3(1), then s contains an even number of b's.

Since:

 L = \{\text{abb}, \text{bab}, \text{bba}\} 

All three string s in L have a length of 3 and all of them have an even number of b's.

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k), that is:

If a string s in L has length 3k, then s contains an even number of b's.

This is the inductive hypothesis.

Prove P(k + 1), that is:

If a string s in L has length 3(k + 1), then s contains an even number of b's.

Now 3(k + 1) = 3k + 3 and the strings in L are obtained by juxtaposing strings already in L with one of abb, bab, or bba. Thus, either the initial or the final three characters in s are abb, bab, or bba. Moreoever, the other 3k characters in s are also in L by definition of L, and so, by the inductive hypothesis, the other 3k characters in s contain an even number, say m, of b's. Because each of abb, bab, and bba contains 2 b's, the total number of b's in s is m + 2, which is a sum of even integers and hence is even.

Q.E.D.

30. A set S consists of strings obtained by juxtaposing one or more copies of 1110 and 0111. Use mathematical induction to prove that for every integer n \geq 1, if a string s in S has a length 4n, then the number of 1's in s is a multiple of 3.

Proof (by mathematical induction):

Suppose a set S contains strings obtained by juxtaposing one or more copies of 1110 and 0111.

Let P(n) be the sentence:

If a string s in S has length 4n, then the number of $1$'s in s is a multiple of 3.

Basis Step:

Prove P(1), that is:

If a string s in S has length 4(1), then the number of $1$'s in s is a multiple of 3.

Since S consists only of strings obtained by juxtaposing 1110 and 0111, then at a minimum, the strings in S must have a length of 4. This means that the only two strings in S that have a length of 4 are 1110 and 0111. The number of $1$'s in s is a multiple of 3 in both cases.

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1.

Suppose P(k), that is:

If a string s in S has length 4k, then the number of $1$'s in s is a multiple of 3.

This is the inductive hypothesis.

Prove P(k + 1), that is:

If a string s in S has length 4(k + 1), then the number of $1$'s in s is a multiple of 3.

Now 4(k + 1) = 4k + 4 and the strings in S are obtained by juxtaposing strings already in S with one of 1110 or 0111. Thus, the number of $1$'s is a multiple of 3 in both cases. Moreover, the other 4k digits in s are also in S by the definition of S, and so, by inductive hypothesis, the other 4k characters in s contain an odd number, say m of $1$'s. Because each of 1110 and 0111 contain 3 $1$'s, the total number of $1$'s in s is m + 1, which is the sum of odd integers and hence is odd.

Q.E.D.

31. Use mathematical induction to give an alternative proof for the statement proved in Example 4.9.9:

For any positive integer n, a complete graph on n vertices has \dfrac{n(n - 1)}{2} edges. Hint: Let P(n) be the sentence, "the number of edges in a complete graph on n vertices is \dfrac{n(n - 1)}{2}."

Omitted.

32. Some 5 \times 5 checkerboards with one square removed can be completely covered by L-shaped trominoes, whereas other 5 \times 5 checkerboards cannot. Find examples of both kinds of checkerboards. Justify your answers.

Omitted.

33. Consider a 4 \times 6 checkerboard. Draw a covering of the board by L-shaped trominoes.

Omitted.

a. Use mathematical induction to prove that for each integer n \geq 1, any checkerboard with dimensions 2 \times 3n can be completely covered with L-shaped trominoes.

Omitted.

b. Let n be any integer greater than or equal to 1. Use the result of part (a) to prove by mathematical induction that for every integer m, any checkerboard with dimensions 2m \times 3n can be completely covered with L-shaped trominoes.

Omitted.

35. Let m and n be any integers that are greater than or equal to 1.

a. Prove that a necessary condition for an m \times n checkerboard to be completely coverable by L-shaped trominoes is that mn be divisible by 3.

Omitted.

b. Prove that having be divisible by 3 is not a sufficient condition for an m \times n checkerboard to be completely covered by L-shaped trominoes.

Omitted.

36. In a round-robin tournament each team plays every other team exactly once with ties not allowed. If the teams are labeled T_1, T_2, \dots, T_n, then the outcome of such a tournament can be represented by a directed graph, in which the teams are represented as dots and an arrow is drawn from one dot to another if, and only if, the following team represented by the first dot beats the team represented by the second dot. For example, the following directed graph shows one outcome of a round-robin tournament involving five teams, A, B, C, D, and E.

See Page 322 for image.

Use mathematical induction to show that in any round-robin tournament involving n teams, where n \geq 2, it is possible to label the teams T_1, T_2, \dots, T_n so that T_i beats T_{i + 1} for all i = 1, 2, \dots n - 1,. (For instance, one such labeling in the example above is T_1 = 1, T_2 = B, T_3 = C, T_4 = E, T_5 = D.) (Hint: Given k + 1 teams, pick one - say T' - and apply the inductive hypothesis to the remaining teams to obtain an ordering T_1, T_2, \dots, T_k. Consider three cases: T' beats T_1, T' loses to the first m teams (where 1 \leq m \leq k - 1) and beats the $(m + 1)$st team, and T' loses to all the other teams.)

Omitted.

37. On the outside rim of a circular disk the integers from 1 through 30 are painted in random order. Show that no matter what this order is, there must be three successive integers whose sum is at least 45.

Omitted.

38. Suppose that n a's and n b's are distributed around the outside of a circle. Use mathematical induction to prove that for any integer n \geq 1, given any such arrangement, it is possible to find a starting point so that if you travel around the circle in a clock-wise direction, the number of a's you pass is never less than the number of b's you have passed. For example, in the diagram shown below, you could start at the a with an asterisk.

See Page 322 for image.

Omitted.

39. For a polygon to be convex means that given any two points on or inside the polygon, the line joining the points lies entirely inside the polygon. Use mathematical induction to prove that for every integer n \geq 3, the angles of any $n$-sided convex polygon add up to 180(n - 2) degrees.

Omitted.

a. Prove that in an 8 \times 8 checkerboard with alternating black and white squares, if the squares in the top right and bottom left corners are removed the remaining board cannot be covered with dominoes. (Hint: Mathematical induction is not needed for this proof.)

Omitted.

b. Use mathematical induction to prove that for each positive integer n, if a 2n \times 2n checkerboard with alternating black and white squares has one white square and one black square removed anywhere on the board, the remaining squares can be covered with dominoes.

Omitted.

41. A group of people are positioned so that the distance between any two people is different from the distance between any other two people. Suppose that the group contains an odd number of people and each person sends a message to their nearest neighbor. Use mathematical induction to prove that at least one person does not receive a message from anyone. [This exercise is inspired by the article "Odd Pie Fights" by L. Carmony, The Mathematics Teacher, 72(1), 1979, 61-64.]

Omitted.

42. Show that for any integer n, it is possible to find a group of n people who are all positioned so that the distance between any two people is different from the distance between any other two people, so that each person sends a message to their nearest neighbor, and so that every person in the group receives a message from another person in the group.

Omitted.

43. Define a game as follows: You begin with an urn that contains a mixture of white and black balls, and during the game you have access to as many additional white and black balls as you might need. In each move you remove two balls from the urn without looking at their colors. If the balls are the same color, you put in one black ball. If the balls are different colors, you put the white ball back into the urn and keep the black ball out. Because each move reduces the number of balls in the urn by one, the game will end with a single ball in the urn. If you know how many white balls and how many black balls are initially in the urn, can you predict the color of the ball at the end of the game? [This exercise is based on one described in "Why correctness must be a mathematical concern" by E.W. Djikstra, www.cs.utexas.edu/users/EWD/transcriptions/EWD07xx/EWD720.html.]

a. Map out all possibilities for playing the game starting with two balls in the urn, then three balls, and then four balls. For each case keep track of the number of white and black balls you start with and the color of the ball at the end of the game.

Omitted.

b. Does the number of white balls seem to be predictive? Does the number of black balls seem to be predictive? Make a conjecture about the color of the ball at the end of the game given the numbers of white and black balls at the beginning.

Omitted.

c. Use mathematical induction to prove the conjecture you made in part (b).

Omitted.

44. Let P(n) be the following sentence: Given any graph G with n vertices satisfying the condition that every vertex of G has degree at most M, then the vertices of G can be colored with at most M + 1 colors in such a way that no two adjacent vertices have the same color. Use mathematical induction to prove this statement is true for every integer n \geq 1.

In order for a proof by mathematical induction to be valid, the basis statement must be true for n = a and the argument of the inductive step must be correct for every integer k \geq a. IN 45 and 46 find the mistakes in the "proofs" by mathematical induction.

Omitted.

"Theorem:" For any integer n \geq 1, all the numbers in a set of n numbers are equal to each other.

"Proof (by mathematical induction): It is obviously true that all the numbers in a set consisting of just one number are equal to each other, so the basis step is true. For the inductive step, let A = \{a_1, a_2, \dots, a_k, a_{k + 1}\} be any set of k + 1 numbers. Form two subsets each of size k:

 B = \{a_1, a_2, a_3, \dots, a_k\} \text{ and } 

 C = \{a_1, a_3, a_4, \dots, a_{k + 1}} 

(B consists of all the numbers in A except a_{k + 1}, and C consists of all the numbers in A except a_2.) By inductive hypothesis, all the numbers in B equal a_1 and all the numbers in C equal a_1 (since both sets have only k numbers). But every number in A is in B or C, so all the numbers in A equal a_1; hence all are equal to each other."

Omitted.

"Theorem:" For every integer n \geq 1, 3^n - 2 is even.

"Proof (by mathematical induction): Suppose the theorem is true for an integer k, where k \geq 1. That is, suppose that 3^k - 2 is even. We must show that 3^{k + 1} - 2 is even. Observe that

 3^{k + 1} - 2 = 3^k \cdot 3 - 2 = 3^k(1 + 2) - 2 

 = (3^k - 2) + 3^k \cdot 2 

Now 3^k - 2 is even by inductive hypothesis and 3^k \cdot 2 is even by inspection. Hence the sum of the two quantities is even (by Theorem 4.1.1). It follows that 3^{k + 1} - 2 is even, which is what we needed to show."

Omitted.

---

Exercise Set 5.4

Page 333

1. Suppose a_1, a_2, a_3, \dots is a sequence defined as follows:

 a_1 = 1, a_2 = 3, a_k = a_{k - 2} + 2a_{k - 1} 

for each integer k \geq 3.

Prove that a_n is odd for every integer n \geq 1.

Proof (by strong mathematical induction):

Let the property P(n) be the sentence "a_n is odd."rim

Basis Step:

Prove P(1) and P(2) are true. That is:

 a_1 \text{ is odd} 

and

 a_2 \text{ is odd} 

Observe from the given definition of the sequence that a_1 = 1, which means that P(1) is true since 1 is odd. Also, observe that a_2 = 3, which means that P(2) is true since 3 is odd.

Inductive Step:

Let k be any integer with k \geq 2. Suppose a_i is odd for each integer i with 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1) is true.

By the definition of the sequence, we know that

 a_{k + 1} = a_{k - 1} + 2a_k 

By the inductive hypothesis, a_{k - 1} is odd.

Also, every term in the sequence is an integer by the sum of products of integers, and so 2a_k is even by the definition of even. It follows that a_{k + 1} is the sum of an odd integer and an even integer. By Theorem 4.1.2, the sum of an odd and even integer is odd. Therefore a_{k + 1} is odd, and P(k + 1) is true.

Q.E.D.

2. Suppose b_1, b_2, b_3, \dots is a sequence defined as follows:

 b_1 = 4, b_2 = 12, b_k = b_{k - 2} + b_{k - 1} 

for each integer k \geq 3.

Prove that b_n is divisible by 4 for every integer n \geq 1.

Proof (by strong mathematical induction):

Let P(n) be the sentence "b_n is divisible by 4."

Basis Step:

Prove P(1) and P(2). That is:

 b_1 \text{ is divisible by } 4 

and

 b_2 \text{ is divisible by } 4 

By the given sequence, we know that b_1 = 4, which is divisible by 4 since 4 = 4 \cdot 1. Also b_2 = 12, which is divisible by 4 since 12 = 4 \cdot 3. Therefore P(1) and P(2) are true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose that b_i is divisible by 4 for each integer 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 b_{k + 1} \text{ is divisible by } 4 

By the definition of the sequence, we know that

 b_{k + 1} = b_{k - 1} + b_k 

By the inductive hypothesis, we know that b_{k - 1} and b_k are both divisible by 4. By the definition of divisibility, b_{k + 1} can be represented as follows:

 b_{k + 1} = 4r + 4s 

where r and s are some integers. By algebra then:

 b_{k + 1} = 4(r + s) 

Now, r + s is an integer by the sum of integers. By the definition of divisibility, b_{k + 1} is divisible by 4. Therefore P(k + 1) is true.

Q.E.D.

3. Suppose that c_0, c_1, c_2, \dots is a sequence defined as follows:

 c_0 = 2, c_1 = 2, c_2 = 6, c_k = 3c_{k - 3} 

for every integer k \geq 3.

Prove that c_n is even for each integer n \geq 0.

Proof (by strong mathematical induction):

Let P(n) be the sentence "c_n is even."

Basis Step:

Prove P(0), P(1), and P(2). That is:

 c_0 \text{ is even} 

and

 c_1 \text{ is even} 

and

 c_2 \text{ is even} 

By the given sequence c_0 = 2, and 2 is even by the definition of even. Also, c_1 = 2, and 2 is even by the definition of even. Also, c_2 = 6, and 6 is even by the definition of even. Therefore P(0), P(1), and P(2) are true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose c_i is even for each integer i with 0 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 c_{k + 1} \text{ is even} 

By the given sequence, we know that:

 c_{k + 1} = 3c_{k - 2} 

By the inductive hypothesis, we know that c_{k - 2} is even. c_{k + 1} can then be represented as:

 c_{k + 1} = 3(2r) 

for some integer r.

Then, by algebra:

 c_{k + 1} = 6r 

 c_{k + 1} = 2(3r) 

Now, 3r is an integer by the product of integers. It follows that c_{k + 1} is even by the definition of even. Therefore P(k + 1) is true.

Q.E.D.

4. Suppose that d_1, d_2, d_3, \dots is sequence defined as follows:

 d_1 = \frac{9}{10}, d_2 = \frac{10}{11}, d_k = d_{k - 1} \cdot d_{k - 2} 

for every integer k \geq 3.

Prove that 0 < d_n \leq 1 for each integer n \geq 1.

Proof (by strong mathematical induction):

Let P(n) be the sentence "0 < d_n \leq 1."

Basis Step:

Prove P(1) and P(2). That is:

 0 < d_1 \leq 1 

and

 0 < d_2 \leq 1 

By the given sequence we know that d_1 = \dfrac{9}{10}, and that 0 < \dfrac{9}{10} \leq 1. Also, we know that d_2 = \dfrac{10}{11}, and that 0 < \dfrac{10}{11} \leq 1. Therefore P(1) and P(2) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose 0 < d_i \leq 1 for each integer i with 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 0 < d_{k + 1} \leq 1 

By the given sequence, we know that:

 d_{k + 1} = d_k \cdot d_{k - 1} 

By the inductive hypothesis, we know that 0 < d_k \leq 1 and that 0 < d_{k - 1} \leq 1. Consequently, 0 < d_{k + 1} \leq 1 because the product of two positive numbers less than or equal to 1 is itself less than or equal to 1. Therefore P(k + 1) is true.

Q.E.D.

5. Suppose that e_0, e_1, e_2, \dots is a sequence defined as follows:

 e_0 = 12, e_1 = 29, e_k, = 5e_{k - 1} - 6e_{k - 2} 

for each integer k \geq 2.

Prove that e_n = 5 \cdot 3^n + 7 \cdot 2^n for every integer n \geq 0.

Proof (by strong mathematical induction):

Let P(n) be the sentence "e_n = 5 \cdot 3^n + 7 \cdot 2^n."

Basis Step:

Prove P(0) and P(1). That is:

 e_0 = 5 \cdot 3^0 + 7 \cdot 2^0 

and

 e_1 = 5 \cdot 3^1 + 7 \cdot 2^1 

By the given sequence, we know that e_0 = 12. By algebra/arithmetic:

 12 = 5 \cdot 3^0 + 7 \cdot 2^0 

 12 = 5 \cdot 1 + 7 \cdot 1 

 12 = 5 + 7 

 12 = 12 

By the given sequence, we know that e_1 = 29. By algebra/arithmetic:

 29 = 5 \cdot 3^1 + 7 \cdot 2^1 

 29 = 5 \cdot 3 + 7 \cdot 2 

 29 = 15 + 14 

 29 = 29 

Therefore P(0) and P(1) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose e_i = 5 \cdot 3^i + 7 \cdot 2^i for each integer i with 0 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 e_{k + 1} = 5 \cdot 3^{k + 1} + 7 \cdot 2^{k + 1} 

By the given sequence, we know that:

 e_{k + 1} = 5e_{k} - 6e_{k - 1} 

By the inductive hypothesis and substitution, e_{k + 1} can be rewritten as:

 e_{k + 1} = 5(5 \cdot 3^k + 7 \cdot 2^k) - 6(5 \cdot 3^{k - 1} + 7 \cdot 2^{k - 1}) 

 = 25 \cdot 3^k + 35 \cdot 2^k - 30 \cdot 3^{k - 1} - 42 \cdot 2^{k - 1} 

 = 25 \cdot 3^k + 35 \cdot 2^k - 10 \cdot 3 \cdot 3^{k - 1} - 21 \cdot 2 \cdot 2^{k - 1} 

 = 25 \cdot 3^k + 35 \cdot 2^k - 10 \cdot 3^k - 21 \cdot 2^k 

 = (25 - 10) \cdot 3^k + (35 - 21) \cdot 2^k 

 = 15 \cdot 3^k + 14 \cdot 2^k 

 = 5 \cdot 3 \cdot 3^k + 7 \cdot 2 \cdot 2^k 

 = 5 \cdot 3^{k + 1} + 7 \cdot 2^{k + 1} 

Therefore P(k + 1) is true.

Q.E.D.

6. Suppose that f_0, f_1, f_2, \dots is a sequence defined as follows:

 f_0 = 5, f_1 = 16, f_k = 7f_{k - 1} - 10f_{k - 2} 

for every integer k \geq 2.

Prove that f_n = 3 \cdot 2^n + 2 \cdot 5^n for each integer n \geq 0.

Proof (by strong mathematical induction):

Let P(n) be the sentence "f_n = 3 \cdot 2^n + 2 \cdot 5^n."

Basis Step:

Prove P(0) and P(1). That is:

 f_0 = 3 \cdot 2^0 + 2 \cdot 5^0 

 f_1 = 3 \cdot 2^1 + 2 \cdot 5^1 

By the given sequence, we know that f_0 = 5. So, by algebra/arithmetic:

 5 = 3 \cdot 2^0 + 2 \cdot 5^0 

 5 = 3 \cdot 1 + 2 \cdot 1 

 5 = 3 + 2 

 5 = 5 

By the given sequence, we know that f_1 = 16. So, by algebra/arithmetic:

 16 = 3 \cdot 2^1 + 2 \cdot 5^1 

 16 = 3 \cdot 2 + 2 \cdot 5 

 16 = 6 + 10 

 16 = 16 

Therefore P(0) and P(1) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose f_i = 3 \cdot 2^i + 2 \cdot 5^i for each integer i with 0 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 f_{k + 1} = 3 \cdot 2^{k + 1} + 2 \cdot 5^{k + 1} 

By the given sequence, we know that:

 f_{k + 1} = 7f_k - 10f_{k - 1} 

By the inductive hypothesis and substitution, f_{k + 1} can be rewritten as:

 f_{k + 1} = 7(3 \cdot 2^k + 2 \cdot 5^k) - 10(3 \cdot 2^{k - 1} + 2 \cdot 5^{k - 1}) 

 = (21 \cdot 2^k + 14 \cdot 5^k) - (30 \cdot 2^{k - 1} + 20 \cdot 5^{k - 1}) 

 = (21 \cdot 2 \cdot 2^{k - 1} + 14 \cdot 5 \cdot 5^{k - 1}) - (30 \cdot 2^{k - 1} + 20 \cdot 5^{k - 1}) 

 = 42 \cdot 2^{k - 1} + 70 \cdot 5^{k - 1} - 30 \cdot 2^{k - 1} - 20 \cdot 5^{k - 1} 

 = (42 - 30) \cdot 2^{k - 1} + (70 - 20) \cdot 5^{k - 1} 

 = 12 \cdot 2^{k - 1} + 50 \cdot 5^{k - 1} 

 = 3 \cdot 2 \cdot 2 \cdot 2^{k - 1} + 2 \cdot 5 \cdot 5 \cdot 5^{k - 1} 

 = 3 \cdot 2^{k + 1} + 2 \cdot 5^{k + 1} 

Therefore P(k + 1) is true.

Q.E.D.

7. Suppose that g_1, g_2, g_3, \dots is a sequence defined as follows:

 g_1 = 3, g_2 = 5, g_k = 3g_{k - 1} - 2g_{k - 2} 

for each integer k \geq 3.

Prove that g_n = 2^n + 1 for every integer n \geq 1.

Proof (by strong mathematical induction):

Let P(n) be the sentence "g_n = 2^n + 1."

Basis Step:

Prove P(1) and P(2). That is:

 g_1 = 2^1 + 1 

and

 g_2 = 2^2 + 1 

By the given sequence, we know that g_1 = 3. By algebra/arithmetic:

 3 = 2^1 + 1 

 3 = 2 + 1 

 3 = 3 

By the given sequence, we know that g_2 = 5. By algebra/arithmetic:

 5 = 2^2 + 1 

 5 = 4 + 1 

 5 = 5 

Therefore P(1) and P(2) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose g_i = 2^i + 1 for each integer i with 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 g_{k + 1} = 2^{k + 1} + 1 

By the given sequence, we know that:

 g_{k + 1} = 3g_k - 2g_{k - 1} 

By the inductive hypothesis and substitution, g_{k + 1} can be rewritten as:

 g_{k + 1} = 3(2^k + 1) - 2(2^{k - 1} + 1) 

 = 3 \cdot 2^k + 3 - 2 \cdot 2^{k - 1} - 2 

 = 3 \cdot 2 \cdot 2^{k - 1} + 3 - 2 \cdot 2^{k - 1} - 2 

 = 6 \cdot 2^{k - 1} + 3 - 2 \cdot 2^{k - 1} - 2 

 = (6 - 2) \cdot 2^{k - 1} + 3 - 2 

 = 4 \cdot 2^{k - 1} + 1 

 = 2 \cdot 2 \cdot 2^{k - 1} + 1 

 = 2 \cdot 2^{k} + 1 

 = 2^{k + 1} + 1 

Therefore P(k + 1) is true.

Q.E.D.

8. Suppose that h_0, h_1, h_2, \dots is a sequence defined as follows:

 h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k - 1} + h_{k - 2} + h_{k - 3} 

for each integer k \geq 3.

a. Prove that h_n \leq 3^n for every integer n \geq 0.

Proof (by strong mathematical induction):

Let P(n) be the sentence "h_n \leq 3^n."

Basis Step:

Prove P(0), P(1), and P(2). That is:

 h_0 \leq 3^0 

and

 h_1 \leq 3^1 

and

 h_2 \leq 3^2 

By the given sequence we know that h_0 = 1. By substitution:

 1 \leq 3^0 

 1 \leq 1 

By the given sequence we know that h_1 = 2. By substitution:

 2 \leq 3^1 

 2 \leq 3 

By the given sequence we know that h_2 = 3. By substitution:

 3 \leq 3^2 

 3 \leq 9 

Therefore P(0), P(1), and P(2) are all true.

Inductive Step:

Let k be any integer where k \geq 3. Suppose h_i \leq 3^i for each integer i with 0 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 h_{k + 1} \leq 3^{k + 1} 

By the given sequence, we know that:

 h_{k + 1} = h_k + h_{k - 1} + h_{k - 2} 

 = h_k + h_{k - 1} + h_{k - 2} \leq 3^k + 3^{k - 1} + 3^{k - 2} 

 = h_k + h_{k - 1} + h_{k - 2} \leq 3^{k - 2}(3^2 + 3^1 + 1) 

 = h_k + h_{k - 1} + h_{k - 2} \leq 3^{k - 2}(9 + 3 + 1) 

 = h_k + h_{k - 1} + h_{k - 2} \leq 13 \cdot 3^{k - 2} 

Since 3^{k + 1} = 3^3 \cdot 3^{k - 2} = 27 \cdot 3^{k - 2}, we know then that:

 = h_k + h_{k - 1} + h_{k - 2} \leq 13 \cdot 3^{k - 2} \leq 27 \cdot 3^{k - 2} = 3^{k + 1} 

Therefore P(k + 1) is true.

Q.E.D.

b. Suppose that s is any real number such that s^3 \geq s^2 + s + 1. (This implies that 2 > s > 1.83.) Prove that h_n \leq s^n for every integer n \geq 2.

Omitted.

9. Define a sequence a_1, a_2, a_3, \dots as follows: a_1 = 1, a_2 = 3, and a_k = a_{k - 1} + a_{k - 2} for every integer k \geq 3. (This sequence is known as the Lucas sequence.) Use strong mathematical induction to prove that a_n \leq \left(\dfrac{7}{4}\right)^n for every integer n \geq 1.

Proof (by strong mathematical induction):

Let P(n) be the sentence "a_n \leq \left(\dfrac{7}{4}\right)^n."

Basis Step:

Prove P(1) and P(2). That is:

 a_1 \leq \left(\dfrac{7}{4}\right)^1 

and

 a_2 \leq \left(\dfrac{7}{4}\right)^2 

By the given sequence, we know that a_1 = 1. By substitution:

 1 \leq \left(\dfrac{7}{4}\right)^1 

 1 \leq \dfrac{7}{4} = 1.75 

By the given sequence, we know that a_2 = 3. By substitution:

 3 \leq \left(\dfrac{7}{4}\right)^2 

 3 \leq \dfrac{49}{16} = 3.0625 

Therefore P(1) and P(2) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose a_i \leq \left(\dfrac{7}{4}\right)^i for each integer i with 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

 a_{k + 1} \leq \left(\dfrac{7}{4}\right)^{k + 1} 

By the given sequence, we know that:

 a_{k + 1} = a_k + a_{k - 1} 

 = a_k + a_{k - 1} \leq \left(\frac{7}{4}\right)^k + \left(\frac{7}{4}\right)^{k - 1} 

 \leq \left(\frac{7}{4}\right) \cdot \left(\frac{7}{4}\right)^{k - 1} + \left(\frac{7}{4}\right)^{k - 1} 

 \leq \left(\frac{7}{4} + 1\right) \cdot \left(\frac{7}{4}\right)^{k - 1} 

 \leq \left(\frac{11}{4}\right) \cdot \left(\frac{7}{4}\right)^{k - 1} 

Since we know that:

 \left(\dfrac{7}{4}\right)^{k + 1} = \frac{7}{4} \cdot \frac{7}{4} \cdot \left(\frac{7}{4}\right)^{k - 1} 

 \left(\dfrac{7}{4}\right)^{k + 1} = \frac{49}{16} \cdot \left(\frac{7}{4}\right)^{k - 1} 

Since \dfrac{11}{4} < \dfrac{49}{16}, it follows that:

 a_{k + 1} = \frac{11}{4} \cdot \left(\frac{7}{4}\right)^{k - 1} \leq \frac{49}{16} \cdot \left(\frac{7}{4}\right)^{k - 1} = \left(\dfrac{7}{4}\right)^{k + 1} 

Therefore P(k + 1) is true.

Q.E.D.

10. The introductory example solved with ordinary mathematical induction in Section 5.3 can also be solved using strong mathematical induction. Let P(n) be "any $n$¢ can be obtained using a combination of $3$¢ and $5$¢ coins." Use strong mathematical induction to prove that P(n) is true for every integer n \geq 8.

Proof (by strong mathematical induction):

Let P(n) be the sentence "any $n$¢ can be obtained using a combination of $3$¢ and $5$¢ coins."

Basis Step:

Prove P(8) and P(9).

P(8) is true because $8$¢ can be obtained by using one $3$¢ coin and one $5$¢ coin.

P(9) is true because $9$¢ can be obtained by using three $3$¢ coins.

Therefore P(8) and P(9) are both true.

Inductive Step:

Let k be any integer where k \geq 8. Suppose P(i) is true for every integer i where 8 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

"any $(k + 1)$¢ can be obtained using a combination of $3$¢ and $5$¢ coins."

Case 1 (There is a $5$¢ coin among those that used to make up the $k$¢.):

In this case replace the $5$¢ coin by two $3$¢ coins; the result will be $(k + 1)$¢.

Case 2 (There is not a $5$¢ coin among those used to make up the $k$¢.):

In this case, because k \geq 8, at least three $3$¢ coins must have been used. So remove three $3$¢ coins and replace them by two $5$¢ coins; the result will be $(k + 1)$¢.

Thus in either case $(k + 1)$¢ can be obtained using $3$¢ and $5$¢ coins.

Q.E.D.

11. You begin solving a jigsaw puzzle by finding two pieces that match and fitting them together. Every subsequent step of the solution consists of fitting together two blocks, each of which is made up of one or more pieces that have previously been assembled. Use strong mathematical induction to prove that for every integer n \geq 1, the number of steps required to put together all n pieces of a jigsaw puzzle is n - 1.

Proof (by strong mathematical induction):

Let P(n) be the sentence "For every integer n \geq 1, the number of steps required to put together all n pieces of a jigsaw puzzle is n - 1."

Basis Step:

Prove P(1). That is:

"For every integer 1 \geq 1, the number of steps required to put together all 1 pieces of a jigsaw puzzle is 1 - 1 = 0."

Since there is only 1 piece of the jigsaw puzzle, it follows that it takes 0 steps to complete the jigsaw puzzle.

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose that P(i) is true, where 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

"For every integer (k + 1) \geq 1, the number of steps required to put together all (k + 1) pieces of a jigsaw puzzle is (k + 1) - 1 = k."

Consider assembling a jigsaw puzzle consisting of k + 1 pieces. The last step involves fitting together two blocks. Suppose one of the blocks consists of r pieces and the other consists of s pieces (where r and s are some integers.) Then r + s = k + 1 and 1 \leq r \leq k and 1 \leq s \leq k.

By the inductive hypothesis, the number of steps required to assemble the blocks are r - 1 and s - 1, respectively.

Then, the total number of steps required to assemble the puzzle is (r - 1) + (s - 1) + 1 = (r + s) - 1 = (k + 1) - 1 = k.

Therefore P(k + 1) is true.

Q.E.D.

12. The sides of a circular track contain a sequence of n cans of gasoline. For each integer n \geq 1, the total amount in the cans is sufficient to enable a certain car to make one complete circuit of the track. In addition, all the gasoline could fit into the car's gas tank at one time. Use mathematical induction to prove that it is possible to find an initial location for placing the car so that it will be able to traverse the entire track by using the various amounts of gasoline in the cans that it encounters along the way.

Proof (by strong mathematical induction):

Let P(n) be the sentence:

For any circular track containing n gasoline cans whose total gasoline is enough for one complete circuit (and all gasoline fits in the tank), there exists an initial location at which the car can start and successfully traverse the entire track.

Basis Step:

Prove P(1). That is:

For any circular track containing 1 gasoline cans whose total gasoline is enough for one complete circuit (and all gasoline fits in the tank), there exists an initial location at which the car can start and successfully traverse the entire track.

It follows from the given problem statement that since there is 1 gasoline can whose total gasoline is enough for one complete circuit, that the initial location at which the car can start and successfully traverse the entire track is the location of this 1 gasoline can.

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) is true for every integer i where 1 \leq i \leq k.

This is the inductive hypothesis.

Prove P(k + 1). That is:

For any circular track containing (k + 1) gasoline cans whose total gasoline is enough for one complete circuit (and all gasoline fits in the tank), there exists an initial location at which the car can start and successfully traverse the entire track.

Consider an arbitrary circular track with k + 1 gasoline cans. Since the total amount of gasoline in the cans is sufficient to enable the car to make one complete circuit of the track, at least one gasoline can must contain enough gasoline to enable the car to travel to the next can.

Take such a can and transfer its gasoline to the can immediately preceding it in the direction of travel. This reduces the number of cans from k + 1 to k.

By the inductive hypothesis, the resulting configuration with k cans can be traversed starting from some initial location. This starting location also works for the k + 1 can configuration, since the redistribution of gasoline does not prevent traversal of the track.

Q.E.D.

13. Use strong mathematical induction to prove the existence part of the unique factorization of integers theorem (Theorem 4.4.5). In other words, prove that every integer greater than 1 is either a prime number or a product of prime numbers.

Proof (by strong mathematical induction):

Let P(n) be the sentence "n is either a prime number or a product of prime numbers."

Basis Step:

Prove P(2). That is:

"2 is either a prime number or a product of prime numbers."

By the definition of prime numbers, 2 is a prime number.

Therefore P(2) is true.

Inductive Step:

Let k be any integer where k > 1. Suppose P(i), for every i where 2 \leq i \leq k, that is:

"i is either a prime number or a product of prime numbers."

Prove P(k + 1). That is:

"(k + 1) is either a prime number or a product of prime numbers."

Case where (k + 1) is prime:

Since (k + 1) is prime, P(k + 1) is true.

Case where (k + 1) is composite (not prime):

Since (k + 1) is composite, this means that k + 1 can be written as:

 k + 1 = a \cdot b 

where a and b are some integers such that 2 \leq a \leq k and 2 \leq b \leq k.

By the inductive hypothesis, this means that both P(a) and P(b) are true. It follows then that a \cdot b is a product of primes and that k + 1 is a product of primes. Therefore P(k + 1) is true.

Q.E.D.

14. Any product of two or more integers is a result of successive multiplications of two integers at a time. For instance, here are a few of the ways in which a_1a_2a_3a_4 might be computed: (a_1a_2)(a_3a_4) or (((a_1a_2)a_3)a_4) or a_1((a_2a_3)a_4). Use strong mathematical induction to prove that any product of two or more odd integers is odd.

Proof (by strong mathematical induction):

Let P(n) be the sentence "any product of n \geq 2 odd integers is odd."

Basis Step:

Prove P(2). That is:

"any product of 2 odd integers is odd."

The following is a proof from 4.2 (exercise 20) that proves this:

Suppose n is any odd integer and m is any odd integer.

Since n and m are odd integers, n = 2k + 1 and m = 2s + 1 where k is some integer and s is some integer.

Then:

 n \cdot m = (2k + 1)(2s + 1) \quad \text{ by substitution} 

 \quad = 4ks + 2s + 2k + 1 

 \quad = 2(2ks + s + k) + 1  \quad \text{ by algebra} 

Let t = 2ks + s + k.

Then n \cdot m = 2(2ks + s + k) + 1 = 2t + 1 where t is an integer because the products and sums of integers is an integer.

Therefore n \cdot m is odd by the definition of odd integers and P(2) is true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose P(i) for every integer i where 2 \leq i \leq k. That is:

"any product of i odd integers is odd."

This is the inductive hypothesis.

Prove P(k + 1). That is:

"any product of (k + 1) odd integers is odd."

Consider the product of a series of odd integers up until k + 1 integers:

 [a_1 \cdot a_2 \cdot a_3 \cdot \dots \cdot a_k] \cdot a_{k + 1} 

By the inductive hypothesis we know that [a_1 \cdot a_2 \cdot a_3 \cdot \dots \cdot a_k] is odd. Thus, we can rewrite this as:

 (2r + 1) \cdot a_{k + 1} 

where r is some integer.

Now 2r + 1 is an integer by the sum and product of integers and 2r + 1 is odd by the definition of odd. The product of (2r + 1) \cdot a_{k + 1} is odd by the proof provided in the basis step. Thus the product of k + 1 odd integers is odd.

Therefore P(k + 1) is true.

Q.E.D.

15. Define the "sum" of one integer to be that integer, and use strong mathematical induction to prove that for every integer n \geq 1, any sum of n even integers is even.

Proof (by strong mathematical induction):

Let P(n) be the sentence "any sum of n even integers is even."

Basis Step:

Prove P(1). That is:

"any sum of 1 even integers is even."

Let r be any even integer. Since r is even, r = 2s for some integer s.

By the problem statement, the sum of one integer is that integer. Therefore the sum of r is r, which is even.

Therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) is true for every integer i where 1 \leq i \leq k. That is:

"any sum of i even integers is even."

This is the inductive hypothesis.

Prove P(k + 1). That is:

"any sum of (k + 1) even integers is even."

Consider a series of even integers up until k + 1 integers:

 a_1, a_2, a_3, \dots, a_{k + 1} 

Now consider the sum of these even integers:

 a_1 + a_2 + a_3 + \dots + a_{k + 1} 

This can also be written as:

 [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} 

By the inductive hypothesis we know that [a_1 + a_2 + a_3 + \dots + a_k] is even. Then we can rewrite this sum as:

 (2q) + a_{k + 1} 

for some integer q.

Also, since we know that a_{k + 1} is even, we can further rewrite this as:

 (2q) + (2u) 

for some integer u.

Then this becomes, by algebra:

 2(q + u) 

Now q + u is an integer by the sum of integers, and 2(q + u) is even by the definition of even. Thus, the sum of k + 1 integers is even.

Therefore P(k + 1) is true.

Q.E.D.

16. Use strong mathematical induction to prove that for every integer n \geq 2, if n is even, then any sum of n odd integers is even, and if n is odd, then any sum of n odd integers is odd.

Proof (by strong mathematical induction):

Let P(n) be the sentence "If n is even, then any sum of n odd integers is even, and if n is odd, then any sum of n odd integers is odd."

Basis Step:

Prove P(2) and P(3).

For P(2):

"If 2 is even, then any sum of 2 odd integers is even, and if 2 is odd, then any sum of 2 odd integers is odd."

Since 2 is even:

Let m and p be any 2 odd integers. Since both m and p are odd, m = 2q + 1 and p = 2r + 1 for some integers q and r.

Their sum then is:

 m + p = 2q + 1 + 2r + 1 

 = 2q + 2r + 2 

 = 2(q + r + 1) 

Now q + r + 1 is an integer by the sum of integers. Also, 2(q + r + 1) is even by the definition of even. Thus P(2) is true.

and

For P(3):

"If 3 is even, then any sum of 3 odd integers is even, and if 3 is odd, then any sum of 3 odd integers is odd."

Since 3 is odd:

Let a, b, and c be any 3 odd integers. Since a, b, and c are odd, then a = 2z + 1, b = 2y + 1, and c = 2x + 1, for some integers z, y, and x.

Their sum then is:

 a + b + c = (2z + 1) + (2y + 1) + (2x + 1) 

 = 2z + 2y + 2x + 2 + 1 

 = 2(z + y + x + 1) + 1 

Now, z + y + x + 1 is an integer by the sum of integers, and 2(z + y + x + 1) + 1 is odd by the definition of odd. Thus P(3) is true.

Therefore P(2) and P(3) are both true.

Inductive Step:

Let k be any integer where k \geq 2. Suppose P(i) for every integer i where 2 \leq i \leq k. That is:

"If i is even, then any sum of i odd integers is even, and if i is odd, then any sum of i odd integers is odd."

Prove P(k + 1). That is:

"If (k + 1) is even, then any sum of (k + 1) odd integers is even, and if (k + 1) is odd, then any sum of (k + 1) odd integers is odd."

Case (k + 1) is odd:

Consider a series of odd integers up until k + 1 integers:

 a_1, a_2, a_3, \dots, a_{k + 1} 

Their sum would be:

 a_1 + a_2 + a_3 + \dots + a_{k + 1} 

Alternatively:

 [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} 

By the definition of odd, if k + 1 is odd, then k is even. By the inductive hypothesis then, we know that [a_1 + a_2 + a_3 + \dots + a_k] is even. Thus, we can rewrite our summation as:

 2r + a_{k + 1} 

for some integer r.

Since we know that a_{k + 1} is odd, we can further rewrite our summation as:

 2r + (2s + 1) 

for some integer s.

Then, by algebra:

 2(r + s) + 1 

Now, r + s is an integer by the sum of integers, and 2(r + s) + 1 is odd by the definition of odd.

Thus P(k + 1) is true in this case.

Case (k + 1) is even:

Consider a series of odd integers up until k + 1 integers:

 a_1, a_2, a_3, \dots, a_{k + 1} 

Their sum would be:

 a_1 + a_2 + a_3 + \dots + a_{k + 1} 

Alternatively:

 [a_1 + a_2 + a_3 + \dots + a_k] + a_{k + 1} 

By the definition of even, if k + 1 is even, then k is odd. By the inductive hypothesis then, we know that [a_1 + a_2 + a_3 + \dots + a_k] is odd. Thus, we can rewrite our summation as:

 (2r + 1) + a_{k + 1} 

for some integer r.

Since we know that a_{k + 1} is odd, we can further rewrite our summation as:

 (2r + 1) + (2s + 1) 

for some integer s.

Then, by algebra:

 2r + 2s + 2 

 2(r + s + 1) 

Now, r + s + 1 is an integer by the sum of integers, and 2(r + s + 1) is even by the definition of even. Thus P(k + 1) is true in this case.

Therefore in both cases P(k + 1) is true.

Q.E.D.

17. Compute 4^1, 4^2, 4^3, 4^4, 4^5, 4^6, 4^7, and 4^8. Make a conjecture about the units digit of 4^n where n is a positive integer. Use strong mathematical induction to prove your conjecture.

4^1 = 4 \ 4^2 = 16 \ 4^3 = 64 \ 4^4 = 256 \ 4^5 = 1024 \ 4^6 = 4096 \ 4^7 = 16384 \ 4^8 = 65536 \

Conjecture:

For some integer n \geq 1, if n is odd, then the units digit of 4^n is 4, if n is even, then the units digit of 4^n is 6.

Proof (by strong mathematical induction):

Let P(n) be the sentence: "the units digit of 4^n is 4 if n is odd and 6 if n is even."

Basis Step:

Prove P(1) and P(2).

For P(1), since 1 is odd, then the units of digit of 4^1 should be 4. Evaluating 4^1:

 4^1 = 4 

The units digit of 4^1 is 4, so P(1) is true.

For P(2), since 2, is even, then the units digit of 4^2 should be 6. Evaluating 4^2:

 4^2 = 16 

The units digit of 4^2 is 6, so P(2) is true.

Therefore both P(1) and P(2) are true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) for every integer i where 1 \leq i \leq k. That is:

"the units digit of 4^i is 4 if i is odd and 6 if i is even."

Prove P(k + 1). That is:

"the units digit of 4^{k + 1} is 4 if (k + 1) is odd and 6 if (k + 1) is even."

Case (k + 1) is even:

Consider the following:

 4^{k + 1} = 4 \cdot 4^k 

By the definition of even, if k + 1 is even, then k is odd. Thus 4^k is 4 to the power of an odd integer. By the inductive hypothesis, we know that this means that:

 4^{k + 1} = 4 \cdot (10m + 4) 

for some integer m.

By algebra:

 = 40m + 16 

 = 10(4m + 1) + 6 

Where 4m + 1 is an integer by the sum and product of integers. Thus the units digit of 4^{k + 1} is 6.

Therefore P(k + 1) is true in this case.

Case (k + 1) is odd:

Consider the following:

 4^{k + 1} = 4 \cdot 4^k 

By the definition of odd, if k + 1 is odd, then k is even. Thus 4^k is 4 to the power of an even integer. By the inductive hypothesis, we know that this means that:

 4^{k + 1} = 4 \cdot (10m + 6) 

for some integer m.

By algebra:

 = 40m + 24 

 = 10(4m + 2) + 4 

Where 4m + 2 is an integer by the sum and product of integers. Thus the units digit of 4^{k + 1} is 4.

Therefore P(k + 1) is true in this case.

Therefore, in both cases P(k + 1) is true.

Q.E.D.

18. Compute 9^0, 9^1, 9^2, 9^3, 9^4, and 9^5. Make a conjecture about the units digit of 9^n where n is a positive integer. Use strong mathematical induction to prove your conjecture.

9^0 = 1 \ 9^1 = 9 \ 9^2 = 81 \ 9^3 = 729 \ 9^4 = 6561 \ 9^5 = 59049 \

Conjecture:

For any integer n \geq 0, the units digit of 9^n is 1 if n is even, and 9 if n is odd.

Proof (by strong induction):

Let P(n) be the sentence: "the units digit of 9^n is 1 if n is even, and 9 if n is odd."

Basis Step:

Prove P(0) and P(1).

For P(0):

Since 0 is even, the units digit of 9^0 is claimed to be 1. Evaluate 9^0:

 9^0 = 1 

Thus P(0) is true.

For P(1):

Since 1 is odd, the units digit of 9^1 is claimed to be 9. Evaluate 9^1;

 9^1 = 9 

Thus P(1) is true.

Therefore both P(0) and P(1) are true.

Inductive Step:

Let k be any integer where k \geq 0. Suppose P(i) for every integer i where 0 \leq i \leq k. That is:

"the units digit of 9^i is 1 if i is even, and 9 if i is odd."

Prove P(k + 1). That is:

"the units digit of 9^{k + 1} is 1 if (k + 1) is even, and 9 if (k + 1) is odd."

Case where (k + 1) is even:

Consider:

 9^{k + 1} = 9 \cdot 9^k 

By the definition of even, if k + 1 is even, then k is odd.

By the inductive hypothesis, we know that the units digit of 9^k is 9 if k is odd. We can then rewrite 9^{k + 1} as:

 9^{k + 1} = 9 \cdot (10m + 9) 

for some integer m.

Then, by algebra:

 9^{k + 1} = 90m + 81 

 = 10(9m + 8) + 1 

Where 9m + 8 is an integer by the sum and product of integers. Thus the units digit of 9^{k + 1} is 1.

Therefore P(k + 1) is true in this case.

Case where (k + 1) is odd:

Consider:

 9^{k + 1} = 9 \cdot 9^k 

By the definition of odd, if k + 1 is odd, then k is even.

By the inductive hypothesis, we know that the units digit of 9^k is 1 if k is even. We can then rewrite 9^{k + 1} as:

 9^{k + 1} = 9 \cdot (10m + 1) 

for some integer m.

Then, by algebra:

 9^{k + 1} = 90m + 9 

 = 10(9m) + 9 

Where 9m is an integer by the product of integers. Thus the units digit of 9^{k + 1} is 9.

Therefore P(k + 1) is true in this case.

Therefore P(k + 1) is true in all cases.

Q.E.D.

19. Suppose that a_1, a_2, a_3, \dots is a sequence defined as follows:

a_1 = 1 a_k = 2 \cdot a_{\lfloor \frac{k}{2} \rfloor}

for every integer k \geq 2.

Prove that a_n \leq n for each integer n \geq 1.

Proof (by strong mathematical induction):

Let a_1, a_2, a_3 \dots be a sequence that satisfies the recurrence relation a_k = 2 \cdot a_{\lfloor \frac{k}{2} \rfloor} for every integer k \geq 2, with the initial condition a_1 = 1.

Let P(n) be the inequality a_n \leq n.

Basis Step:

Prove P(1). By the given sequence, we know that a_1 = 1. Then:

 1 \leq 1 

This is a true statement, therefore P(1) is true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) for every integer i where 1 \leq i \leq k. That is:

 a_k \leq k 

Prove P(k + 1). That is:

 a_{k + 1} \leq (k + 1) 

By the given sequence, we know that:

 a_{k + 1} = 2 \cdot a_{\lfloor \frac{k + 1}{2} \rfloor} 

 \leq 2 \cdot \lfloor \frac{k + 1}{2} \rfloor 

By the inductive hypothesis:

\leq \begin{cases} 2 \cdot \left(\frac{k + 1}{2}\right) & \text{if } k \text{ is odd} \ 2 \cdot \left(\frac{k}{2}\right) & \text{if } k \text{ is even} \end{cases}

\leq \begin{cases} k + 1 & \text{if } k \text{ is odd} \ k & \text{if } k \text{ is even} \end{cases}

 \leq k + 1 

In both cases a_{k + 1} \leq (k + 1). Therefore P(k + 1) is true.

Q.E.D.

20. Suppose that b_1, b_2, b_3, \dots is a sequence defined as follows:

b_1 = 0, b_2 = 3, b_k = 5 \cdot b_{\frac{k}{2}} + 6

for every integer k \geq 3.

Prove that b_n is divisible by 3 for each integer n \geq 1.

Proof (by strong mathematical induction):

Let the sequence, b_1, b_2, b_3, \dots be the sequence that satisfies the recurrence relation b_k = 5 \cdot b_{\lfloor \frac{k}{2} \rfloor} + 6 for every integer k \geq 3, with the initial conditions b_1 = 0 and b_2 = 3.

Let P(n) be the sentence "b_n is divisible by $3$" where n \geq 1.

Basis Step:

Prove P(1) and P(2).

For P(1):

By the given sequence b_1 = 0, and 0 is divisible by 3 since 0 = 0 \cdot 3.

For P(2):

By the given sequence b_2 = 3, and 3 is divisible by 3 since 3 = 1 \cdot 3.

Therefore both P(1) and P(2) are true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) for every integer i such that 1 \leq i \leq k. That is:

"b_i is divisible by $3$"

Prove P(k + 1). That is:

"b_{k + 1} is divisible by $3$"

By the given sequence, we know that:

 b_{k + 1} = 5 \cdot b_{\lfloor \frac{k + 1}{2} \rfloor} + 6 

Since 1 \leq \lfloor \dfrac{k + 1}{2} \rfloor \leq k, then, by the inductive hypothesis, b_{\lfloor \frac{k + 1}{2} \rfloor} is divisible by 3.

By the definition of divisibility, we can then rewrite b_{k + 1} as:

 b_{k + 1} = 5 \cdot 3m + 6 

for some integer m.

Then, by algebra:

 = 15m + 6 

 = 3(5m + 2) 

Now, 5m + 2 is an integer by the sum and product of integers. Thus 3 \mid b_{k + 1}.

Therefore P(k + 1) is true.

Q.E.D.

21. Suppose that c_1, c_2, c_3, \dots is a sequence defined as follows:

 c_0 = 1, c_1 = 1, c_k = c_{\lfloor \frac{k}{2} \rfloor} + c_{\lfloor \frac{k}{2} \rfloor} 

for every integer k \geq 2.

Prove that c_n = n for each integer n \geq 1.

Proof (by strong mathematical induction):

Let the sequence, c_0, c_1, c_2, \dots be the sequence that satisfies the recurrence relation c_k = c_{\lfloor \frac{k}{2} \rfloor} + c_{\lfloor \frac{k}{2} \rfloor} for every integer k \geq 2, with the initial conditions c_0 = 1 and c_1 = 1.

Let P(n) be the equality c_n = n for each integer n \geq 1.

Basis Step:

Prove P(1) and P(2).

For P(1):

Based on the given sequence, we know that c_1 = 1. Thus 1 = 1 is a true statement.

For P(2):

Based on the given recurrence relation:

 c_2 = c_{\lfloor \frac{2}{2} \rfloor} + c_{\lfloor \frac{2}{2} \rfloor} 

 c_2 = c_1 + c_1 

Based on the given sequence, we know that c_1 = 1. By substitution:

 c_2 = 1 + 1 

 c_2 = 2 

2 = 2 is a true statement.

Therefore P(1) and P(2) are both true.

Inductive Step:

Let k be any integer where k \geq 1. Suppose P(i) for every integer i such that 1 \leq i \leq k. That is:

 c_i = i 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 c_{k + 1} = k + 1 

By the given sequence, we know that:

 c_{k + 1} = c_{\lfloor \frac{k + 1}{2} \rfloor} + c_{\lfloor \frac{k + 1}{2} \rfloor} 

Since we know that 1 \leq \lfloor \dfrac{k + 1}{2} \rfloor \leq k, by the inductive hypothesis, we then know that c_{\lfloor \frac{k + 1}{2} \rfloor} = \dfrac{k + 1}{2}. By substitution:

 c_{k + 1} = \frac{k + 1}{2} + \frac{k + 1}{2} 

 = 2\left(\frac{k + 1}{2}\right) 

 = k + 1 

Therefore P(k + 1) is true.

Q.E.D.

22. One version of the game NIM starts with two piles of objects such as coins, stones, or matchsticks. In each turn a player is required to remove from one to three objects from one of the piles. The two players take turns doing this until both piles are empty. The loser is the first player who can't make a move. Use strong mathematical induction to show that if both piles contain the same number of objects at the start of the game, the player who goes second can always win.

Omitted.

23. Define a game G as follows: Begin with a pile of n stones and 0 points. In the first move split the pile into two possibly unequal sub-piles, multiply the number of stones in one sub-pile times the number of stones in the other sub-pile, and add the product to your score. In the second move, split each of the newly created piles into a pair of possibly unequal sub-piles, multiply the number of stones in each sub-pile times the number of stones in the paired sub-pile, and add the new products to your score. Continue by successively splitting each newly created pile of stones that has at least two stones into a pair of sub-piles, multiplying the number of stones in each sub-pile times the number of stones in the paired sub-pile, and adding the new products to your score. The game G ends when no pile contains more than one stone.

a. Play G starting with 10 stones and using the following initial moves. In move 1 split the pile of 10 stones into two sub-piles with 3 and 7 stones respectively, compute 3 \cdot 7 = 21, and find that your score is 21. In move 2 split the pile of 3 stones into two sub-piles, with 1 and 2 stones respectively, and split the pile of 7 stones into two sub-piles, with 4 and 3 stones respectively, compute 1 \cdot 2 = 2 and 4 \cdot 3 = 12, and find that your score is 21 + 2 + 12 = 35. In move 3 split the pile of 4 stones into two sub-piles, each with 2 stones, and split the pile of 3 tones into two sub-piles, with 1 and 2 stones respectively, and find your new score. Continue splitting piles and computing your score until no pile has more than one stone. Show your final score along with a record of the numbers of stones in the piles you created with your moves.

Omitted.

b. Play G again starting with 10 stones, but use a different initial move from the one in part (a). Show your final score along with a record of the numbers of stones in the piles you created with your moves.

Omitted.

c. Show that you can use strong mathematical induction to prove that for every integer n \geq 1, given the set-up of game G, no matter how you split the piles in the various moves, your final score is \dfrac{n(n - 1)}{2}. The basis step may look a little strange because a pile consisting of one stone cannot be split into any sub-piles. Another way to say this is that it can only be split into zero piles, and that gives an answer that agrees with the general formula for the final score.

Omitted.

24. Imagine a situation in which eight people, numbered consecutively 1-8, are arranged in a circle. Starting from person #1, every second person in the circle is eliminated. The elimination process continues until only one person remains. In the first round the people numbered 2, 4, 6, and 8 are eliminated, in the second round the people numbered 3 and 7 are eliminated, and in the third round person #5 is eliminated, so after the third round only person #1 remains, as shown on the next page.

See page 336 for image.

a. Given a set of sixteen people arranged in a circle and numbered, consecutively 1-16, list the numbers of the people who are eliminated in each round if every second person is eliminated and the elimination process continues until only one person remains. Assume that the starting point is person #1.

Omitted.

b. Use ordinary mathematical induction to prove that for every integer n \geq 1, given any set of 2^n people arranged in a circle and numbered consecutively 1 through 2^n, if one starts from person #1 and goes repeatedly around the circle successively eliminating every second person, eventually only person #1 will remain.

Omitted.

c. Use the result of part (b) to prove that for any nonnegative integers n and m with 2^n \leq 2^n + m < 2^{n + 1}, if r = 2^n + m, then given any set of r people arranged in a circle and numbered consecutively 1 through r, if one starts from person #1 and goes repeatedly around the circle successively eliminating every second person, eventually only person #(2m + 1) will remain.

Omitted.

25. Find the mistake in the following "proof" that purports to show that every nonnegative integer power of every nonzero real number is 1.

"Proof:

Let r be any nonzero real number and let the property P(n) be the equation r^n = 1.

Show that P(0) is true:

P(0) is true because r^0 = 1 by definition of zeroth power.

Show that for every integer k \geq 0, if P(i) is true for each integer i from 0 through k, then P(k + 1) is also true:

Let k be any integer k \geq 0 and suppose that r^i = 1 for each integer i from 0 through k. This is the inductive hypothesis.

We must show that r^{k + 1} = 1. Now

 r^{k + 1} = r^{k + k - (k - 1)} 

because k + k - (k - 1) = k + k - k + 1 = k + 1

 = \frac{r^k \cdot r^k}{r^{k - 1}} 

by the laws of exponents

 = \frac{1 \cdot 1}{1} 

by inductive hypothesis

 = 1 

Thus r^{k + 1} = 1 [as was to be shown].

[Since we have proved both the basis and the inductive step of the strong mathematical induction, we conclude that the given statement is true.]"

Omitted.

26. Use the well-ordering principle for the integers to prove Theorem 4.4.4: Every integer greater than 1 is divisible by a prime number.

Omitted.

27. Use the well-ordering principle for the integers to prove the existence part of the unique factorization of integers theorem. In other words, prove that every integer greater than 1 is either prime or a product of prime numbers.

Omitted.

a. The Archimedean property for the rational numbers states that for every rational number r, there is an integer n such that n > r. Prove this property.

Omitted.

b. Prove that given any rational number r, the number -r is also rational.

Omitted.

c. Use the results of parts (a) and (b) to prove that given any rational number r, there is an integer m such that m < r.

Omitted.

29. Use the results of exercise 28 and the well-ordering principle for the integers to show that given any rational number r, there is an integer m such that m \leq r < m + 1.

Omitted.

30. Use the well-ordering principle to prove that given any integer n \geq 1, there exists an odd integer m and a nonnegative integer k such that n = 2^k \cdot m.

Omitted.

31. Give examples to illustrate the proof of Theorem 5.4.1.

Omitted.

32. Suppose P(n) is a property such that

    1. P(0), P(1), P(2) are all true,

Omitted.

2. for each integer $k \geq 0$, if $P(k)$ is true, then $P(3k)$ is true.
   Must it follow that $P(n)$ is true for every integer $n \geq 0$? If yes,
   explain why; if no, give a counterexample.

Omitted.

33. Prove that if a statement can be proved by strong mathematical induction, then it can be proved by ordinary mathematical induction. To do this, let P(n) be a property that is defined for each integer n, and suppose the following two statements are true:

    1. P(a), P(a + 1), P(a + 2) \dots, P(b).

Omitted.

2. For any integer $k \geq b$, if $P(i)$ is true for each integer $i$ from
   $a$ through $k$, then $P(k + 1)$ is true.

Omitted.

The principle of strong mathematical induction would allow us to conclude immediately that P(n) is true for every integer n \geq a. Can we reach the same conclusion using the principle of ordinary mathematical induction? Yes! To see this, let Q(n) be the property

P(j) is true for each integer j with a \leq j \leq n.

Then use ordinary mathematical induction to show that Q(n) is true for every integer n \geq b. That is, prove:

1. $Q(b)$ is true.

Omitted.

2. For each integer $k \geq b$, if $Q(k)$ is true then $Q(k + 1)$ is true.

Omitted.

34. It is a fact that every integer n \geq 1 can be written in the form

 c_r \cdot 3^r + c_{r - 1} \cdot 3^{r - 1} + \dots + c_2 \cdot 3^2 + c_1 \cdot 3 + c_0 

where c_r = 1 or 2 and c_i = 0, 1, or 2 for each integer i = 0, 1, 2, \dots, r - 1. Sketch a proof of this fact.

Omitted.

35. Use mathematical induction to prove the existence part of the quotient-remainder theorem. In other words, use mathematical induction to prove that given any integer n and any positive integer d, there exists integers q and r such that n = dq + r and 0 \leq r < d.

Omitted.

36. Prove that if a statement can be proved using ordinary mathematical induction, then it can be proved by the well-ordering principle.

Omitted.

37. Use the principle of ordinary mathematical induction to prove the well-ordering principle for the integers.

Omitted.

---

Exercise Set 5.5

Page 346

Exercises 1-5 contain a while loop and a predicate. In each case show that if the predicate is true before entry to the loop, then it is also true after exit from the loop.

loop:

\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := m + 1\\ \ \ \ \ n := n - 1\\ \text{\textbf{end while}}

predicate: m + n = 100

Proof:

Suppose the predicate m + n = 100 is true before entry to the loop. Then

 m_{\text{old}} + n_{\text{old}} = 100 

After execution of the loop,

 m_{\text{new}} = m_{\text{old}} + 1 

and

 n_{\text{new}} = n_{\text{old}} - 1 

so

 m_{\text{new}} + n_{\text{new}} = (m_{\text{old}} + 1) + (n_{\text{old}} - 1) 

 = m_{\text{old}} + n_{\text{old}} = 100 

Q.E.D.

loop:

\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := m + 4\\ \ \ \ \ n := n - 2\\ \text{\textbf{end while}}

predicate: m + n \text{ is odd}

Proof:

Suppose the predicate m + n \text{ is odd} is true before entry to the loop. Then

 m_{\text{old}} + n_{\text{old}} \text{ is odd} 

After execution of the loop,

 m_{\text{new}} = m_{\text{old}} + 4 

and

 n_{\text{new}} = n_{\text{old}} - 2 

so

 m_{\text{new}} + n_{\text{new}} = (m_{\text{old}} + 4) + (n_{\text{old}} - 2) 

 = m_{\text{old}} + n_{\text{old}} + 2 

Since m_{\text{old}} + n_{\text{old}} \text{ is odd}, then:

 = 2k + 1 + 2 

for some integer k

 = 2k + 2 + 1 

 = 2(k + 2) + 1 

Now, k + 2 is an integer by the sum of integers. Therefore m_{\text{new}} + n_{\text{new}} is odd by the definition of odd.

Q.E.D.

loop:

\text{\textbf{while}} (m \geq 0 \text{ and } m \leq 100)\\ \ \ \ \ m := 3 \cdot m\\ \ \ \ \ n := 5 \cdot n\\ \text{\textbf{end while}}

predicate: m^3 > n^2

Proof:

Suppose the predicate m^3 > n^2 is true before entry to the loop. Then

 (m_{\text{old}})^3 > (n_{\text{old}})^2 

After execution of the loop,

 m_{\text{new}} = 3 \cdot m_{\text{old}} 

and

 n_{\text{new}} = 5 \cdot n_{\text{old}} 

so

 (m_{\text{new}})^3 = (3m_{\text{old}})^3 = 27(m_{\text{old}})^3 > 27(n_{\text{old}})^2 

Now, since n_{\text{new}} = 5 \cdot n_{\text{old}}, it follows that \dfrac{1}{5}n_{\text{new}} = n_{\text{old}}. Hence

 (m_{\text{new}})^3 > 27(n_{\text{old}})^2 = 27\left(\frac{1}{5}n_{\text{new}}\right)^2 = 27 \cdot \frac{1}{25}(n_{\text{new}})^2 

 = \frac{27}{25} \cdot (n_{\text{new}})^2 > (n_{\text{new}})^2 

loop:

\text{\textbf{while}} (n \geq 0 \text{ and } n \leq 100)\\ \ \ \ \ n := n + 1\\ \text{\textbf{end while}}

predicate: 2^n < (n + 2)!

Proof:

Suppose the predicate 2^n < (n + 2)! is true before entry to the loop. Then

 2^{n_{\text{old}}} < (n_{\text{old}} + 2)! 

After execution of the loop,

 n_{\text{new}} = n_{\text{old}} + 1 

so

 2^{n_{\text{new}}} = 2^{n_{\text{old}} + 1} = 2 \cdot 2^{n_{\text{old}}} < 2(n_{\text{old}} + 2)! 

Note that 2 \leq n_{\text{old}} + 3

since the guard condition gives n_{\text{old}} \geq 0, then:

 2(n_{\text{old}} + 2)! \leq (n_{\text{old}} + 3)(n_{\text{old}} + 2)! = (n_{\text{old}} + 3)! = ((n_{\text{old}} + 1) + 2)! = (n_{\text{new}} + 2)! 

Combining these gives:

 2^{n_{\text{new}}} = 2 \cdot 2^{n_{\text{old}}} < (n_{\text{old}} + 3)! = (n_{\text{new}} + 2)! 

Q.E.D.

loop:

\text{\textbf{while}} (n \geq 3 \text{ and } n \leq 100)\\ \ \ \ \ n := n + 1\\ \text{\textbf{end while}}

predicate: 2n + 1 \leq 2^n

Proof:

Suppose the predicate 2n + 1 \leq 2^n is true before entry to the loop. Then

 2n_{\text{old}} + 1 \leq 2^{n_{\text{old}}}

After execution of the loop,

 n_{\text{new}} = n_{\text{old}} + 1 

so

 2n_{\text{new}} + 1 = 2(n_{\text{old}} + 1) + 1 = 2n_{\text{old}} + 3 

and

 2^{n_{\text{new}}} = 2^{n_{\text{old}} + 1} = 2 \cdot 2^{n_{\text{old}}} 

If we take the predicate and multiply both sides by 2, we get:

 2(2n_{\text{old}} + 1) \leq 2(2^{n_{\text{old}}}) 

 4n_{\text{old}} + 2 \leq 2^{n_{\text{old}} + 1} = 2^{n_{\text{new}}} 

Notice that the new value for the left-hand value of the inequality is:

 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 

And that this is less than the predicate's left hand side after multiplied by two:

 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 < 4n_{\text{old}} + 2 

And put together this is:

 2n_{\text{new}} + 1 = 2n_{\text{old}} + 3 < 4n_{\text{old}} + 2 \leq 2^{n_{\text{old}} + 1} = 2^{n_{\text{new}}} 

Q.E.D.

Exercises 6-9 each contain a while loop annotated with a pre-and a post-condition and also a loop invariant. In each case, use the loop invariant theorem to prove the correctness of the loop with respect to the pre-and post-conditions.

6. [Pre-condition: m is a nonnegative integer, x is a real number, i = 0, and \text{exp} = 1.]

\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. \text{exp} := \text{exp} \cdot x\\ \ \ \ \ 2. i := i + 1\\ \text{\textbf{end while}}

[Post-condition: $\text{exp} = x^m$]

loop invariant: I(n) is "\text{exp} = x^n and i = n."

I. Basis Property: [I(0) is true before the first iteration of the loop.]

I(0) is "\text{exp} = x^0 and i = 0." According to the pre-condition, before the first iteration of the loop \text{exp} = 1 and i = 0. Since x^0 = 1, I(0) is evidently true.

II. Inductive Property: [If G \wedge I(k) is true before a loop iteration (where k \geq 0), then I(k + 1) is true after the loop iteration.]

Suppose k is any nonnegative integer such that G \wedge I(k) is true before an iteration of the loop. Then as execution reaches the top of the loop i \neq m, \text{exp} = x^k, and i = k. Since i \neq m, the guard is passed and statement 1 is executed. Now before the execution of statement 1,

 \text{exp}_{\text{old}} = x^k 

so execution of statement 1 has the following effect:

 \text{exp}_{\text{new}} = \text{exp}_{\text{old}} \cdot x = x^k \cdot x = x^{k + 1} 

Similarly, before statement 2 is executed,

 i_{\text{old}} = k 

so after execution of statement 2,

 i_{\text{new}} = i_{\text{old}} + 1 = k + 1 

Hence after the loop iteration, the two statements \text{exp} = x^{k + 1} and i = k + 1 are true, and so I(k + 1) is true.

III. Eventual Falsity of Guard: [After a finite number of iterations of the loop, G becomes false.]

The guard G is the condition i \neq m, and m is a nonnegative integer. By I and II, it is known that_

for every integer n \geq 0, if the loop is iterated n times, then \text{exp} = x^n and i = n.

So after m iterations of the loop, i = m. Thus G becomes false after m iterations of the loop.

IV. Correctness of the Post-Condition: [If N is the least number of iterations after which G is false and I(N) is true, then the value of the algorithm variables will be as specified in the post-condition of the loop.]

According to the post-condition, the value of \text{exp} after execution of the loop should be x^m. But when G is false, i = m. And when I(N) is true, i = N and \text{exp} = x^N. Since both conditions (G is false and I(N) is true) are satisfied, m = i = N and \text{exp} = x^m, as required.

7. [Pre-condition: \text{largest} = A[1] and $i = 1$]

\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. i := i + 1\\ \ \ \ \ 2. \text{\textbf{if}} A[i] > \text{largest \textbf{then } \text{largest}} := A[i]\\ \text{\textbf{end while}}

[Post-condition: $\text{largest} = \text{maximum value of } A[1], A[2], \dots, A[m]$]

loop invariant: I(n) is "\text{largest} = \text{maximum value of } A[1], A[2], \dots, A[n + 1] and i = n + 1."

I. Basis Property: [I(0) is true before the first iteration of the loop.]

I(0) is "\text{largest} = A[1] and i = 1." According to the pre-condition, this statement is true.

II. Inductive Property: [If G \wedge I(k) is true before a loop iteration (where k \geq 0), then I(k + 1) is true after the loop iteration.]

Suppose k is any nonnegative integer such that G \wedge I(k) is true before an iteration of the loop. Then as execution reaches the top of the loop, i \neq m, \text{largest} = A[k + 1] and i = k + 1. Since i \neq m, the guard is passed and statement 1 is executed. Now, before execution of statement 1:

 i_{\text{old}} = k + 1 

so after statement 1 is executed:

 i_{\text{new}} = i_{\text{old}} + 1 = k + 2 

Also, before statement 2 is executed:

 \text{largest}_{\text{old}} = \max(A[1], \dots, A[k + 1]) 

so after statement 2 is executed:

\text{largest}{\text{new}} = \begin{cases} A[k + 2] & \text{if } A[k + 2] > \text{largest}{\text{old}} \ \text{largest}{\text{old}} & \text{if } A[k + 2] \leq \text{largest}{\text{old}} \end{cases}

Thus, after the loop iteration, I(k + 1) is true.

III. Eventual Falsity of Guard: [After a finite number of iterations of the loop, G becomes false.]

The guard G is the condition i \neq m. By I and II, it is known that for every integer n \geq 1, after n iterations of the loop, I(n) is true. Hence after m - 1 iterations of the loop, i = m and G is false.

IV. Correctness of the Post-Condition: [If N is the least number of iterations after which G is false and I(N) is true, then the value of the algorithm variables will be as specified in the post-condition of the loop.]

Suppose that N is the least number of iterations after which G is false and I(N) is true. Then (since G is false) i = m and (since I(N) is true) i = N + 1 and \text{largest} = \max(A[1], \dots A[N + 1]). Putting these together gives m = N + 1, and so \text{largest} = \max(A[1], \dots A[m]), which is the post-condition.

Q.E.D.

8. [Pre-condition: \text{sum} = A[1] and $i = 1$]

\text{\textbf{while}} (i \neq m)\\ \ \ \ \ 1. i := i + 1\\ \ \ \ \ 2. \text{sum} := \text{sum} + A[i]\\ \text{\textbf{end while}}

[Post condition: $\text{sum} = A[1] + A[2] + \dots + A[m]$]

loop invariant: I(n) is "i = n + 1 and \text{sum} = A[1] + A[2] + \dots + A[n + 1]."

I. Basis Property: [I(0) is true before the first iteration of the loop.]

I(0) is "i = 1 and \text{sum} = A[1]." According to the pre-condition, this statement is true.

II. Inductive Property: [If G \wedge I(k) is true before a loop iteration (where k \geq 0), then I(k + 1) is true after the loop iteration.]

Suppose k is a nonnegative integer such that G \wedge I(k) is true before the iteration of the loop. Then as execution reaches the top of the loop, i \neq m, i = k + 1, and \text{sum} = A[1] + A[2] + \dots + A[k + 1]. Since i \neq m, the guard is passed and statement 1 is executed. Now before execution of statement 1, i_{\text{old}} = k + 1. So after execution of statement 1, i_{\text{new}} = i_{\text{old}} + 1 = (k + 1) + 1 = k + 2. Also before statement 2 is executed \text{sum}_{\text{old}} = A[1] + A[2] + \dotts + A[k + 1]. Execution of statement 2 adds A[k + 2] to this sum, and so after statement 2 is executed, \text{sum}_{\text{new}} = A[1] + A[2] + \dots + A[k + 1] + A[k + 2]. Thus after the loop iteration, I(k + 1) is true.

III. Eventual Falsity of Guard: [After a finite number of iterations of the loop, G becomes false.]

The guard is the condition i \neq m. By I and II, it is known that for every integer n \geq 1, after n iterations of the loop I(n) is true. Hence, after m - 1 iterations of the loop, I(m) is true, which implies that i = m and G is false.

IV. Correctness of the Post-Condition: [If N is the least number of iterations after which G is false and I(N) is true, then the value of the algorithm variables will be as specified in the post-condition of the loop.]

Suppose that N is the least number of iterations after which G is false and I(N) is true. Then (since G is false) i = m and (since I(N) is true) i = N + 1 and \text{sum} = A[1] + A[2] + \dots + A[N + 1]. Putting these together gives m = N + 1, and so \text{sum} = A[1] + A[2] + \dots A[m], which is the post-condition.

Q.E.D.

9. [Pre-condition: a = A and A is a positive integer.]

\text{\textbf{while}} (a > 0)\\ \ \ \ \ a := a - 2\\ \text{\textbf{end while}}

[Post-condition: a = 0 if A is even and a = -1 if A is odd.]

loop invariant: I(n) is "Both a and A are even integers or both are odd integers and, in either case, a \geq -1."

I. Basis Property: [I(0) is true before the first iteration of the loop.]

I(0) is "a = A and A \text{ is a positive integer}." According to the pre-condition, this statement is true.

II. Inductive Property: [If G \wedge I(k) is true before a loop iteration (where k \geq 0), then I(k + 1) is true after the loop iteration.]

Suppose k is a nonnegative integer such that G \wedge I(k) is true before the iteration of the loop. Then as execution reaches the top of the loop, a_{\text{old}} > 0, a_{\text{old}} \text{ has the same parity as } A, and a_{\text{old}} \geq -1.

Since a_{\text{old}} > 0, it follows that a_{\text{old}} \geq 1. The guard condition allows the loop body to execute, and statement 1 is performed. This results in:

 a_{\text{new}} = a_{\text{old}} - 1 

Thus I(k + 1) is true.

III. Eventual Falsity of Guard: [After a finite number of iterations of the loop, G becomes false.]

The guard is the condition a > 0. By I and II, it is known that for every iteration of the loop a := a - 2. Since the initial value of a is A, this means that the value for a follows the following sequence:

 A, A - 2, A - 4, \dots 

which eventually reaches a value at a \leq 0.

Hence, after a finite number of iterations of the loop, a \leq 0 and G is false.

IV. Correctness of the Post-Condition: [If N is the least number of iterations after which G is false and I(N) is true, then the value of the algorithm variables will be as specified in the post-condition of the loop.]

Suppose that N is the least number of iterations after which G is false and I(N) is true. Then (since G is false) a \leq 0 and (since I(N) is true) both a and A have the same parity and a \geq -1. This means that:

 -1 \leq a \leq 0 

Therefore, if A is even, then a = 0, and if A is odd, then a = -1. This fulfills the post-condition.

10. Prove correctness of the while loop of Algorithm 4.10.3 (in exercise 27 of Exercise Set 4.10) with respect to the following pre- and post-conditions:

Pre-condition: A and B are positive integers, a = A, and b = B.

Post-condition: One of a or b is zero and the other is nonzero. Whichever is nonzero equals \text{gcd}(A, B).

Use the loop invariant

I(n)

"(1) a and b are nonnegative integers with \text{gcd}(a, b) = \text{gcd}(A, B),

(2) at most one of a and b equals 0,

(3) 0 \leq a + b \leq A + B - n."

Omitted.

11. The following while loop implements a way to multiply two numbers that was developed by the ancient Egyptians.

[Pre-condition: A and B are positive integers, x = A, y = B, and \text{product} = 0.]

\text{\textbf{while}} (y \neq 0)\\ \ \ \ \ r := y \mod 2\\ \ \ \ \ \text{\textbf{if }} r = 0\\ \ \ \ \ \ \ \ \ \text{\textbf{then do }}\\ \ \ \ \ \ \ \ \ \ \ \ \ x := 2 \cdot x\\ \ \ \ \ \ \ \ \ \ \ \ \ y := y \text{ div } 2\\ \ \ \ \ \ \ \ \ \text{\textbf{end do}}\\ \ \ \ \ \text{\textbf{if }} r = 1\\ \ \ \ \ \ \ \ \ \text{\textbf{then do }}\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{product} := \text{product } + x\\ \ \ \ \ \ \ \ \ \ \ \ \ y := y - 1\\ \ \ \ \ \ \ \ \ \text{\textbf{end do}}\\ \text{\textbf{end while}}

[Post-condition: $\text{product } = A \cdot B$]

a. Make a trace table to show that the algorithm gives the correct answer for multiplying A = 13 \text{ times } B = 18.

Omitted.

b. Prove the correctness of this loop with respect to its pre-and post-conditions by using the loop invariant

I(n): "$xy + \text{ product} = A \cdot B$"

Omitted.

12. The following sentence could be added to the loop invariant for the Euclidean algorithm:

There exist integers u, v, s, and t such that a = uA + vB and b = sA + tB.

a. Show that this sentence is a loop invariant for

\text{\textbf{while}} (b \neq 0)\\ \ \ \ \ r := a \mod b\\ \ \ \ \ a := b\\ \ \ \ \ b := r\\ \text{\textbf{end while}}

Omitted.

b. Show that if initially a = A and b = B, then sentence (5.5.12) is true before the first iteration of the loop.

Omitted.

c. Explain how the correctness proof for the Euclidean algorithm together with the results of (a) and (b) above allow you to conclude that given any integers A and B with A > B \geq 0, there exist integers u and v so that \text{gcd}(A, B) = uA + vB.

Omitted.

d. By actually calculating u, v, s, and t at each stage of execution of the Euclidean algorithm, find integers u and v so that \text{gcd}(330, 156) = 330u + 156v.

Omitted.

---

Exercise Set 5.6

Page 360

Find the first four terms of each of the recursively defined sequences in 1-8.

1. a_k = 2a_{k - 1} + k, for every integer k \geq 2 a_1 = 1

a_1 = 1 \ a_2 = 2a_1 + 2 = 2(1) + 2 = 2 + 2 = 4 \ a_3 = 2a_2 + 3 = 2(4) + 3 = 8 + 3 = 11 \ a_4 = 2a_3 + 4 = 2(11) + 4 = 22 + 4 = 26

2. b_k = b_{k - 1} + 3k, for every integer k \geq 2 b_1 = 1

b_1 = 1 \ b_2 = b_1 + 3(2) = 1 + 6 = 7 \ b_3 = b_2 + 3(3) = 7 + 9 = 16 \ b_4 = b_3 + 3(4) = 16 + 12 = 28

3. c_k = k(c_{k - 1})^2, for every integer k \geq 1 c_0 = 1

c_0 = 1 \ c_1 = 1(c_0)^2 = 1(1)^2 = 1(1) = 1 \ c_2 = 2(c_1)^2 = 2(1)^2 = 2(1) = 2 \ c_3 = 3(c_2)^2 = 3(2)^2 = 3(4) = 12

4. d_k = k(d_{k - 1})^2, for every integer k \geq 1 d_0 = 3

d_0 = 3 \ d_1 = 1(d_0)^2 = 1(3)^2 = 1(9) = 9 \ d_2 = 2(d_1)^2 = 2(9)^2 = 2(81) = 162\ d_3 = 3(d_2)^2 = 3(162)^2 = 3(26244) = 78732

5. s_k = s_{k - 1} + 2s_{k - 2}, for every integer k \geq 2, s_0 = 1, s_1 = 1

s_0 = 1 \ s_1 = 1 \ s_2 = s_1 + 2s_0 = 1 + 2(1) = 1 + 2 = 3 \ s_3 = s_2 + 2(s_1) = 3 + 2(1) = 3 + 2 = 5

6. t_k = t_{k - 1} + 2t_{k - 2}, for every integer k \geq 2 t_0 = -1, t_1 = 2

t_0 = -1 \ t_1 = 2 \ t_2 = t_1 + 2t_0 = 2 + 2(-1) = 2 - 2 = 0 \ t_3 = t_2 + 2t_1 = 0 + 2(2) = 0 + 4 = 4

7. u_k = ku_{k - 1} - u_{k - 2}, for every integer k \geq 3 u_1 = 1, u_2 = 1

u_1 = 1 \ u_2 = 1 \ u_3 = 3(u_2) - u_1 = 3(1) - 1 = 3 - 1 = 2 \ u_4 = 4(u_3) - u_2 = 4(2) - 1 = 8 - 1 = 7

8. v_k = v_{k - 1} + v_{k - 2} + 1, for every integer k \geq 3 v_1 = 1, v_2 = 3

v_1 = 1 \ v_2 = 3 \ v_3 = v_2 + v_1 + 1 = 3 + 1 + 1 = 5 \ v_4 = v_3 + v_2 + 1 = 5 + 3 + 1 = 9

9. Let a_0, a_1, a_2, \dots be defined by the formula a_n = 3n + 1, for every integer n \geq 0. Show that this sequence satisfies the recurrence relation a_k = a_{k - 1} + 3, for every integer k \geq 1.

By definition of a_0, a_1, a_2, \dots for each integer k \geq 1,

 \text{(1)} \quad a_k = 3k + 1 

and

 \text{(2)} \quad a_{k - 1} = 3(k - 1) + 1 

Then a_{k - 1} + 3:

 a_{k - 1} + 3 = (3(k - 1) + 1) + 3 \quad \text{ by substitution of (2)} 

 = 3k - 3 + 1 + 3 

 = 3k + 1 \quad \text{ by basic algebra} 

  = a_k \quad \text{ by substitution of (1)} 

10. Let b_0, b_1, b_2, \dots be defined by the formula b_n = 4^n, for every integer n \geq 0. Show that this sequence satisfies the recurrence relation b_k = 4b_{k - 1}, for every integer k \geq 1.

By definition of b_0, b_1, b_2, \dots for each integer k \geq 1,

 \text{(1)} \quad b_k = 4^k 

and

 \text{(2)} \quad b_{k - 1} = 4^{k - 1} 

Then 4b_{k - 1}:

 4b_{k - 1} = 4(4^{k - 1}) \quad \text{ by substitution of (2)} 

 = 4^k \quad \text{ by the laws of exponents} 

 = b_k \quad \text{ by substitution of (1)} 

11. Let c_0, c_1, c_2, \dots be defined by the formula c_n = 2^n - 1 for every integer n \geq 0. Show that this sequence satisfies the recurrence relation c_k = 2c_{k - 1} + 1 for every integer k \geq 1.

By the definition of c_0, c_1, c_2, \dots for each integer k \geq 1,

 \text{(1)} \quad c_k = 2^k - 1 

and

 \text{(2)} \quad c_{k - 1} = 2^{k - 1} - 1 

Then 2c_{k - 1} + 1:

 2c_{k - 1} + 1 = 2(2^{k - 1} - 1) + 1 \quad \text{ by substitution of (2)} 

 = 2^k - 2 + 1 

 = 2^k - 1 

 = c_k \quad \text{ by substitution of (1)} 

12. Let s_0, s_1, s_2, \dots be defined by the formula s_n = \dfrac{(-1)^n}{n!} for every integer n \geq 0. Show that this sequence satisfies the following recurrence relation for every integer k \geq 1:

 s_k = \frac{-s_{k - 1}}{k} 

By the definition of s_0, s_1, s_2, \dots for each integer k \geq 1,

 \text{(1)} \quad s_k = \frac{(-1)^k}{k!} 

and

 \text{(2)} \quad s_{k - 1} = \frac{(-1)^{k - 1}}{(k - 1)!} 

Then \dfrac{-s_{k - 1}}{k}:

 \frac{-s_{k - 1}}{k} = \frac{-1\left(\dfrac{(-1)^{k - 1}}{(k - 1)!}\right)}{k} \quad \text{ by substitution of (2)} 

 = \frac{\dfrac{(-1)^k}{(k - 1)!}}{k} 

 = \frac{(-1)^k}{k(k - 1)!} 

 = \frac{(-1)^k}{k!} 

 = s_k \quad \text{ by substitution of (1)} 

13. Let t_0, t_1, t_2, \dots be defined by the formula t_n = 2 + n for every integer n \geq 0. Show that this sequence satisfies the following recurrence relation for every integer k \geq 2:

 t_k = 2t_{k - 1} - t_{k - 2} 

By the definition of t_0, t_1, t_2, \dots for each integer k \geq 2,

 \text{(1)} \quad t_k = 2 + k 

and

 \text{(2)} \quad t_{k - 1} = 2 + (k - 1) = 1 + k 

and

 \text{(3)} \quad t_{k - 2} = 2 + (k - 2) = k 

Then 2t_{k - 1} - t_{k - 2}:

 2t_{k - 1} - t_{k - 2} = 2(1 + k) - k \quad \text{by substitution of (2) and (3)} 

 = 2 + 2k - k 

 = 2 + k 

 = t_k \quad \text{ by substitution of (1)} 

14. Let d_0, d_1, d_2, \dots be defined by the formula d_n = 3^n - 2^n for every integer n \geq 0. Show that this sequence satisfies the following recurrence relation for every integer k \geq 2:

 d_k = 5d_{k - 1} - 6d_{k - 2} 

By the definition of d_0, d_1, d_2, \dots for each integer k \geq 2,

 \text{(1)} \quad d_k = 3^k - 2^k 

and

 \text{(2)} \quad d_{k - 1} = 3^{k - 1} - 2^{k - 1} 

and

 \text{(3)} \quad d_{k - 2} = 3^{k - 2} - 2^{k - 2} 

Then 5d_{k - 1} - 6d_{k - 2}:

 5d_{k - 1} - 6d_{k - 2} = 5(3^{k - 1} - 2^{k - 1}) - 6(3^{k - 2} - 2^{k - 2}) \quad \text{ by substitution of (2) and (3)} 

 = 5(3^{k - 1} - 2^{k - 1}) - 6(3^{k - 2} - 2^{k - 2}) 

 = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - 6 \cdot 3^{k - 2} + 6 \cdot 2^{k - 2} 

 = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - (3 \cdot 2) \cdot 3^{k - 2} + (3 \cdot 2) \cdot 2^{k - 2} 

 = 5 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} - 2 \cdot 3^{k - 1} + 3 \cdot 2^{k - 1} 

 = 5 \cdot 3^{k - 1} - 2 \cdot 3^{k - 1} - 5 \cdot 2^{k - 1} + 3 \cdot 2^{k - 1} 

 = 3 \cdot 3^{k - 1} - 2 \cdot 2^{k - 1} 

 = 3^k - 2^k 

 = d_k \quad \text{ by substitution of (1)} 

15. For the sequence of Catalan numbers defined in Example 5.6.4, prove that for each integer n \geq 1,

 C_n = \frac{1}{4n + 2}\binom{2n + 2}{n + 1} 

Hint: Mathematical induction is not needed for the proof. Start with the right-hand side of the equation and use algebra to transform it into the left-hand side of the equation.

Recall that:

 C_n = \frac{1}{n + 1}\binom{2n}{n} 

Then \dfrac{1}{4n + 2}\binom{2n + 2}{n + 1}:

 \dfrac{1}{4n + 2}\binom{2n + 2}{n + 1} = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)!}{(n + 1)!((2n + 2) - (n + 1))!}\right) \quad \text{ by definition of binomial} 

 = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)!}{(n + 1)!(n + 1)!}\right) 

 = \frac{1}{2(2n + 1)}\left(\frac{(2n + 2)(2n + 1)(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) \quad \text{ by definition of factorial} 

 = \frac{1}{\cancel{2(2n + 1)}}\left(\frac{\cancel{2}(n + 1)\cancel{(2n + 1)}(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) 

 = \frac{1}{1}\left(\frac{(n + 1)(2n)!}{(n + 1)(n!)(n + 1)(n!)}\right) 

 = \frac{\cancel{(n + 1)}(2n!)}{\cancel{(n + 1)}(n!)(n + 1)(n!)} 

 = \frac{(2n)!}{(n!)(n + 1)(n!)} 

 = \frac{1}{n + 1}\left(\frac{(2n)!}{(n!)(n!)}\right) 

 = \frac{1}{n + 1}\left(\frac{(2n)!}{(n!)(2n - n)!}\right) 

 = \frac{1}{n + 1}\binom{2n}{n} \quad \text{ by definition of binomial} 

 = C_n \quad \text{ by definition of Catalan} 

16. Use the recurrence relation and values for the Tower of Hanoi sequence m_1, m_2, m_3, \dots discussed in Example 5.6.5 to compute m_7 and m_8.

Recall that:

 m_k = 2m_{k - 1} + 1 \quad \text{ recurrence relation} 

and

 m_1 = 1 \quad \text{ initial conditions} 

In Example 5.6.5, we saw that:

 m_2 = 2m_1 + 1 = 2 \cdot 1 + 1 = 3 

 m_3 = 2m_2 + 1 = 2 \cdot 3 + 1 = 7 

 m_4 = 2m_3 + 1 = 2 \cdot 7 + 1 = 15 

 m_5 = 2m_4 + 1 = 2 \cdot 15 + 1 = 31 

 m_6 = 2m_5 + 1 = 2 \cdot 31 + 1 = 63 

Therefore, continuing the computations for m_7 and m_8:

 m_7 = 2m_6 + 1 = 2 \cdot 63 + 1 = 127 

 m_8 = 2m_7 + 1 = 2 \cdot 127 + 1 = 255 

17. Tower of Hanoi with Adjacency Requirement:

Suppose that in addition to the requirement that they never move a larger disk on top of a smaller one, the priests who move the disks of the Tower of Hanoi are also allowed only to move disks one by one from one pole to an adjacent pole. Assume poles A and C are at the two ends of the row and pole B is in the middle. Let

 a_n = \left[\text{the minimum number of moves needed to transfer a tower of } n \text{ disks from pole } A \text{ to pole } C \right] 

a. Find a_1, a_2, and a_3.

 a_1 = 2 

 a_2 = 2 \text{(moves to move the top disk from pole A to pole C)} 

 +1 \text{(move to move the bottom disk from pole A to pole B)} 

 +2 \text{(moves to move top disk from pole C to pole A)} 

 +1 \text{(move to move the bottom disk from pole B to pole C)} 

 +2 \text{(move to move top disk from pole A to pole C)} 

 = 8 

 a_3 = 8 + 1 + 8 + 1 + 8 = 26 

b. Find a_4.

 a_4 = 26 + 1 + 26 + 1 + 26 = 80 

c. Find a recurrence relation for a_1, a_2, a_3, \dots. Justify your answer.

For every integer k \geq 2,

 a_k = a_{k - 1} \text{(moves to move the top } k - 1 \text{ disks from pole A to pole C)} 

 +1 \text{move to move the bottom disk from pole A to pole B} 

 +a_{k - 1} \text{(moves to move the top disk from pole C to pole A)} 

 +1 \text{(move to move the bottom disks from pole B to pole C)} 

 +a_{k - 1} \text{(moves to move the top disks from pole A to pole C)} 

 = 3a_{k - 1} + 2 

18. Tower of Hanoi with Adjacency Requirement:

Suppose the same situation as in exercise 17. Let

 b_n = \left[\text{the minimum number of moves needed to transfer a tower of } n \text{ disks from pole } A \text{ to pole } B \right] 

a. Find b_1, b_2, and b_3.

 b_1 = 1 

 b_2 = 4 

 b_3 = 13 

b. Find b_4.

 b_4 = 40 

c. Show that b_k = a_{k - 1} + 1 + b_{k - 1} for each integer k \geq 2, where a_1, a_2, a_3, \dots is the sequence defined in exercise 17.

First move the top k - 1 disks from A to C, which takes a minimum of a_{k - 1} moves.

Then move the remaining $k$th disk from A to B, which takes a minimum of 1 move.

Then move the k - 1 disks from C to B, on top of the $k$th disk, which takes a minimum of b_{k - 1} moves. (Moving from A to B is the same as moving from C to B, the same number of moves).

These moves are minimal because, due to the adjacency requirement, the top k - 1 disks (have to be) moved to C first.

Therefore:

 b_k = a_{k - 1} + 1 + b_{k - 1} 

d. Show that b_k \leq 3b_{k - 1} + 1 for each integer k \geq 2.

We need to show a_{k - 1} \leq 2b_{k - 1} by part c. This is true because we can first move k - 1 disks from A to B which takes a minimum of b_{k - 1} moves, and then move them from B to C, which takes a minimum of another b_{k - 1} moves. Doing this results in k - 1 disks being moved from A to C, which takes a minimum of a_{k - 1} moves.

Therefore:

 a_{k - 1} \leq 2b_{k - 1} 

e. Show that b_k = 3b_{k - 1} + 1 for each integer k \geq 2.

Proof (by mathematical induction):

Let P(k) by the equation b_k = 3b_{k - 1} + 1.

Basis Step:

Prove P(2). That is:

 b_2 = 3b_1 + 1 

 4 = 3(1) + 1 \quad \text{ by substitution of part (a)} 

 4 = 4 

Therefore P(2) is true.

Inductive Step:

Suppose P(k) is true where k is any integer such that k \geq 2. That is:

 b_{k} = 3b_{k - 1} + 1 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 b_{k + 1} = 3b_{(k + 1) - 1} + 1 = 3b_k + 1 

We know, by part c, that:

 b_{k + 1} = a_k + 1 + b_k 

And we know that a_k \leq 2b_k by part (d).

We need to show a_k \geq 2b_k.

When moving k disks from A to C, consider the largest disk. Due to the adjacency requirement, it has to move to B first. So the top k- 1 disks must have moved to C before that. Then for the largest disk to finally move from B to C, the top k - 1 disks must have first moved from C to A to get out of the way. In the same way, the top k - 1 disks, on their way from C back to B, must have been moved to B (on top of the largest disk) first, before reaching A (This shows that at some point all the disks are on the middle pole.) This takes a minimum of b_k moves. Then moving all the disks from B to C takes a minimum of b_k moves. Therefore a_k \geq 2b_k.

Thus:

 b_{k + 1} = a_k + 1 + b_k = 2b_k + 1 + b_k = 3b_k + 1 

Q.E.D.

19. Four-Pole Tower of Hanoi:

Suppose that the Tower of Hanoi problem has four poles in a row instead of three. Disks can be transferred one by one from one pole to any other pole, but at no time may a larger disk be placed on top of a smaller disk. Let s_n be the minimum number of moves needed to transfer the entire tower of n disks from the left-most to the right-most pole.

a. Find s_1, s_2, and s_3.

 s_1 = 1, s_2 = 1 + 1 + 1 = 3, s_3 = s_1 + (1 + 1 + 1) + s_1 = 5 

b. Find s_4.

 s_4 = s_2 + (1 + 1 + 1) + s_2 = 9 

c. Show that s_k \leq 2s_{k - 2} + 3 for every integer k \geq 3.

Proof:

Let's label the poles A-B-C-D, from left to right.

First notice that, since there is no adjacency requirement, the number of moves to move A to D is equal to the number of moves from any pole to any other pole. So, moving k disks from A to, say, B, still takes s_k moves.

First move the top k - 2 disks from A to B, in s_{k - 2} moves. Then move the second largest disk from A to C. Then move the largest dis to D. Then move the second largest disk from C to D, on top of the largest. Finally, move k - 2 disks from B to D. This takes s_{k - 2} + 1 + 1 + 1 + s_{k - 2} moves.

This procedure gives us the minimum number of moves, because there is no adjacency requirement and we are taking advantage of the free space in all 4 poles. (Notice that this is faster than moving the top k - 1 disks somewhere else first, then moving the largest disk to D, then moving the k - 2 disks. Similarly it's faster than moving k - 3 disks first, then moving the bottom 3, since there are not enough empty poles to make that efficient.)

20. Tower of Hanoi Poles in a Circle:

Suppose that instead of being lined up in a row, the three poles for the original Tower of Hanoi are placed in a circle. The monks move the disks one by one from one pole to another, but they may only move disks one over in a clockwise direction and they may never move a larger disk on top of a smaller one. Let c_n be the minimum number of moves needed to transfer a pile of n disks from one pole to the next adjacent pole in the clockwise direction.

a. Justify the inequality c_k \leq 4c_{k - 1} + 1 for each integer k \geq 2.

Proof:

Label the poles A, B, C, in clockwise order A \to B \to C \to A.

To move k disks from A to B, first move the top k - 1 disks from A to B (which takes c_{k - 1}), then from B to C (which takes c_{k - 1}), then move the largest disk from A to B (which takes 1 move), then move the k - 1 disks from C to A (which takes c_{k - 1}), then from A to B on top of the largest disk (which takes c_{k - 1}).

So the total moves made are 4c_{k - 1} + 1. This shows that moving k disks from A to B can be accomplished in 4c_{k - 1} + 1 moves, so c_k \leq 4c_{k - 1} + 1.

b. The expression 4c_{k - 1} + 1 is not the minimum number of moves needed to transfer a pile of k disks from one pole to another. Explain, for example, why c_3 \neq 4c_2 + 1.

Proof:

c_2 = \ 1 \text{(move to transfer the top disk from A to B)} \ +1 \text{(move to transfer the top disk from B to C)} \ +1 \text{(move to transfer the bottom disk from A to B)} \ +1 \text{(move to transfer the top disk from C to A)} \ +1 \text{(move to transfer the top disk from A to B)} \

c_3 = \ 1 \text{(move to transfer the top disk from A to B)} \ +1 \text{(move to transfer the top disk from B to C)} \ +1 \text{(move to transfer the middle disk from A to B)} \ +1 \text{(move to transfer the top disk from C to A)} \ +1 \text{(move to transfer the middle disk from B to C)} \ +1 \text{(move to transfer the top disk from A to B)} \ +1 \text{(move to transfer the top disk from B to C)} \

After these 7 steps have been completed, the bottom disk can be transferred from A to B. At that point the top two disks are on C, and a modified version of the initial seven steps can be used to transfer them from C to B. Thus the total number of steps is 7 + 1 + 7 = 15, and 15 < 21 = 4c_2 + 1.

21. Double Tower of Hanoi:

In this variation of the Tower of Hanoi there are three poles in a row and 2n disks, two each of n different sizes, where n is any positive integer. Initially one of the poles contains all the disks placed on top of each other in pairs of decreasing size. Disks are transferred one by one from one pole to another, but at no time may a larger disk be placed on top of a smaller disk. However, a disk may be placed on top of one of the same size. Let t_n be the minimum number of moves needed to transfer a tower of 2n disks from one pole to another.

a. Find t_1 and t_2.

Suppose the poles are labeled A, B, and C such that A \to B \to C.

Let s_1 = \text{small disk 1}, and s_2 = \text{small disk 2}.

t_1 = \ 1 & s_1 \to B \ +1 & s_2 \to B \ = 2

Let m_1 = \text{medium disk 1}, and m_2 = \text{medium disk 2}.

t_2 = 1 & s_1 \to B \ +1 & s_2 \to B \ +1 & s_1 \to C \ +1 & s_2 \to C \ +1 & m_1 \to B \ +1 & m_2 \to B \ +1 & s_1 \to B \ +1 & s_2 \to B \ = 8

b. Find t_3.

Let l_1 = \text{large disk 1}, and l_2 = \text{large disk 2}.

t_3 = 1 & s_1 \to B \ +1 & s_2 \to B \ +1 & s_2 \to C \ +1 & s_1 \to C \ +1 & m_1 \to B \ +1 & m_2 \to B \ +1 & s_1 \to B \ +1 & s_2 \to B \ +1 & s_2 \to A \ +1 & s_1 \to A \ +1 & m_2 \to C \ +1 & m_1 \to C \ +1 & s_1 \to B \ +1 & s_2 \to B \ +1 & s_2 \to C \ +1 & s_1 \to C \ +1 & l_1 \to B \ +1 & l_2 \to B \ +1 & s_1 \to B \ +1 & s_2 \to B \ +1 & s_2 \to A \ +1 & s_1 \to A \ +1 & m_1 \to B \ +1 & m_2 \to B \ +1 & s_1 \to B \ +1 & s_2 \to B \ = 26

c. Find a recurrence relation for t_1, t_2, t_3, \dots.

 t_1 = 2, t_2 = 8, t_3 = 26 

 t_n = 3t_{n - 1} + 2 \quad n \geq 2 

22. Fibonacci Variation:

A single pair of rabbits (male and female) is born at the beginning of a year. Assume the following conditions (which are somewhat more realistic than Fibonacci's):

(1) Rabbit pairs are not fertile during their first months of life but thereafter give birth to four new male/female pairs at the end of every month.

(2) No rabbits die.

a. Let r_n = \text{ the number of pairs of rabbits alive at the end of month } n, for each integer n \geq 1, and let r_0 = 1. Find a recurrence relation for r_0, r_1, r_2, \dots. Justify your answer.

This is similar to example 5.6.6, but instead of giving birth to one new pair, each male/female pair of rabbits gives birth to two new pairs after the first month of life.

Proof:

At r_0 = 1, as there is only 1 pair of rabbits and they are not yet fertile. r_1 = 1, as they are no yet fertile until month 2. At month 2, this pair has four pairs, resulting in five pairs of rabbits. r_2 = 4 + 1 = 5.

The four new pairs can only come from the fertile pairs, which become fertile at n - 2 months, where n \in \mathbb{Z}^+ \wedge n \geq 0. The total of infertile pairs can be calculated simply by looking at r_{n - 1}. Therefore the total number of pairs of rabbits at n months can be expressed by the recurrence relation:

 r_n = 4(r_{n - 2}) + r_{n - 1} 

b. Compute r_0, r_1, r_2, r_3, r_4, r_5, and r_6.

r_0 = 1 \ r_1 = 1 \ r_2 = 4(1) + 1 = 5 \ r_3 = 4(1) + 5 = 9 \ r_4 = 4(5) + 9 = 29 \ r_5 = 4(9) + 29 = 65 \ r_6 = 4(29) + 65 = 181 \

c. How many rabbits will there be at the end of the year?

r_0 = 1 \ r_1 = 1 \ r_2 = 4(1) + 1 = 5 \ r_3 = 4(1) + 5 = 9 \ r_4 = 4(5) + 9 = 29 \ r_5 = 4(9) + 29 = 65 \ r_6 = 4(29) + 65 = 181 \ r_7 = 4(65) + 181 = 441 \ r_8 = 4(181) + 441 = 1165 \ r_9 = 4(441) + 1165 = 2929 \ r_{10} = 4(1165) + 2929 = 7589 \ r_{11} = 4(2929) + 7589 = 19305 \ r_{12} = 4(7589) + 19305 = 49661 \

23. Fibonacci Variation:

A single pair of rabbits (male and female) is born at the beginning of a year. Assume the following conditions:

(1) Rabbit pairs are not fertile during their first two months of life but thereafter give birth to three new male/female pairs at the end of every month.

(2) No rabbits die.

a. Let s_n = \text{ the number of pairs of rabbits alive at the end of month } n, for each integer n \geq 1, and let s_0 = 1. Find a recurrence relation for s_0, s_1, s_2, \dots. Justify your answer.

Proof:

We are given that in the beginning, there is only a single pair of rabbits, so s_0 = 1. Since the rabbits are not fertile for the first two months of life, this means that s_1 = 1 and s_2 = 1. Afterwards which the pair of rabbits is fertile and will give birth to three pairs of rabbits. So s_3 = 3s_0 + s_2.

The amount of given rabbits at n months would be 3 times the rabbits that are fertile, which are any rabbits that exist at n - 3 months (s_{n - 3}) plus the amount of infertile rabbits, which is just s_{n - 1} rabbits. This gives the recurrence relation:

 s_n = 3s_{n - 3} + s_{n - 1} 

b. Compute s_0, s_1, s_2, s_3, s_4, and s_5.

s_0 = 1 \ s_1 = 1 \ s_2 = 1 \ s_3 = 3(1) + (1) = 4 \ s_4 = 3(1) + (4) = 7 \ s_5 = 3(1) + (7) = 10 \

c. How many rabbits will there be at the end of the year?

s_0 = 1 \ s_1 = 1 \ s_2 = 1 \ s_3 = 3(1) + (1) = 4 \ s_4 = 3(1) + (4) = 7 \ s_5 = 3(1) + (7) = 10 \ s_6 = 3(4) + (10) = 22 \ s_7 = 3(7) + (22) = 43 \ s_8 = 3(10) + (43) = 73 \ s_9 = 3(22) + (73) = 139 \ s_{10} = 3(43) + (139) = 268 \ s_{11} = 3(73) + (268) = 487 \ s_{12} = 3(139) + (487) = 904 \

In 24-34, F_0, F_1, F_2, \dots is the Fibonacci sequence.

24. Use the recurrence relation and values for F_0, F_1, F_2, \dots given in Example 5.6.6 to compute F_{13} and F_{14}.

The recurrence relation and values given from Example 5.6.6 are:

 F_k = F_{k - 1} + F{k - 2} \quad \text{ recurrence relation} 

 F_0 = 1, F_1 = 1 \quad \text{ initial conditions} 

Luckily, 5.6.6 also gives us F_2 through F_{12}, so now to calculate F_{13} and F_{14}:

F_2 = F_1 + F_0 = 1 + 1 = 2 \ F_3 = F_2 + F_1 = 2 + 1 = 3 \ F_4 = F_3 + F_2 = 3 + 2 = 5 \ F_5 = F_4 + F_3 = 5 + 3 = 8 \ F_6 = F_5 + F_4 = 8 + 5 = 13 \ F_7 = F_6 + F_5 = 13 + 8 = 21 \ F_8 = F_7 + F_6 = 21 + 13 = 34 \ F_9 = F_8 + F_7 = 34 + 21 = 55 \ F_{10} = F_9 + F_8 = 55 + 34 = 89 \ F_{11} = F_{10} + F_9 = 89 + 55 = 144 \ F_{12} = F_{11} + F_{10} = 144 + 89 = 233 \ F_{13} = F_{12} + F_{11} = 233 + 144 = 377 \ F_{14} = F_{13} + F_{12} = 377 + 233 = 610 \

25. The Fibonacci sequence satisfies the recurrence relation F_k = F_{k - 1} + F_{k - 2}, for every integer k \geq 2.

a. Explain why the following is true:

 F_{k + 1} = F_k + F_{k - 1} \text{ for each integer } k \geq 1 

Each term of the Fibonacci sequence beyond the second equals the sum of the previous two. For any integer k \geq 1, the two terms previous to F_{k + 1} are F_k and F_{k - 1}. Hence for every integer k \geq 1, F_{k + 1} = F_k + F_{k - 1}.

b. Write an equation expressing F_{k + 2} in terms of F_{k + 1} and F_k.

The Fibonacci sequence satisfies the recurrence relation:

 F_{k + 2} = F_{k + 1} + F_k 

for each integer k \geq 0.

c. Write an equation expressing F_{k + 3} in terms of F_{k + 2} and F_{k + 1}.

The Fibonacci sequence satisfies the recurrence relation:

 F_{k + 3} = F_{k + 2} + F_{k + 1} 

for each integer k \geq -1.

26. Prove that F_k = 3F_{k - 3} + 2F_{k - 4} for every integer k \geq 4.

Proof:

Since we are trying to express this in terms of F_k, we must look recursively at the definitions of it's preceding two terms until we see them as expressions of F_{k - 3} and F_{k - 4} instead of F_{k - 1} and F_{k - 2}.

 F_k = F_{k - 1} + F_{k - 2} 

 = (F_{k - 2} + F_{k - 3}) + (F_{k - 3} + F_{k - 4}) 

 = ((F_{k - 3} + F_{k - 4}) + F_{k - 3}) + (F_{k - 3} + F_{k - 4}) 

 = F_{k - 3} + F_{k - 4} + F_{k - 3} + F_{k - 3} + F_{k - 4} 

 = 3F_{k - 3} + 2F_{k - 4} 

27. Prove that F_k^2 - F_{k - 1}^2 = F_kF_{k + 1} - F_{k - 1}F_{k + 1}, for every integer k \geq 1.

The standard Fibonacci sequence from 5.6.6 is:

 F_k = F_{k - 1} + F_{k - 2} 

To find the given equation to be true, we must convert the left-hand side to the right hand-side. Meaning we must express the given Fibonacci recurrence relation in terms of F_{k}, F_{k + 1}, and F_{k - 1}.

 F_k^2 - F_{k - 1}^2 = (F_k - F_{k - 1})(F_k + F_{k - 1}) \quad \text{ by algebra of the difference between two squares} 

 = (F_k - F_{k - 1})F_{k + 1} \quad \text{ by the definition of the Fibonacci sequence} 

 = F_kF_{k + 1} - F_{k - 1}F_{k + 1} \quad \text{ by distribution} 

28. Prove that F_{k + 1}^2 - F_k^2 - F_{k - 1}^2 = 2F_kF_{k - 1}, for each integer k \geq 1.

 F_{k + 1} = F_k + F_{k - 1} 

 F_{k + 1}^2 = (F_k + F_{k - 1})^2 

 = (F_k + F_{k - 1})(F_k + F_{k - 1}) 

 = F_k^2 + 2F_{k}F_{k - 1} + F_{k - 1}^2 

 (F_{k + 1}^2) - F_k^2 - F_{k - 1}^2 = (F_k^2 + 2F_{k}F_{k - 1} + F_{k - 1}^2) - F_k^2 - F_{k - 1}^2 

 = 2F_kF_{k - 1} 

29. Prove that F_{k + 1}^2 - F_k^2 = F_{k - 1}F_{k + 2}, for every integer k \geq 1.

 F_{k + 1} = F_k + F_{k - 1} 

 F_{k + 1}^2 = (F_k + F_{k - 1})^2 

 = (F_k + F_{k - 1})(F_k + F_{k - 1}) 

 = F_k^2 + 2F_kF_{k - 1} + F_{k - 1}^2 

 (F_{k + 1}^2) - F_k^2 = (F_k^2 + 2F_kF_{k - 1} + F_{k - 1}^2 ) - F_k^2 

 = 2F_kF_{k - 1} + F_{k - 1}^2 

 = F_{k - 1}(2F_k + F_{k - 1}) 

 = F_{k - 1}(F_{k - 1} + F_k + F_k) 

 = F_{k - 1}((F_k + F_{k - 1}) + F_k) 

 = F_{k - 1}((F_{k + 1}) + F_k) 

 = F_{k - 1}(F_{k + 1} + F_k) 

 = F_{k - 1}(F_{k + 2}) 

 = F_{k - 1}F_{k + 2} 

30. Use mathematical induction to prove that for each integer n \geq 0, F_{n + 2}F_n - F_{n + 1}^2 = (-1)^n.

Proof (by mathematical induction):

Let P(n) be the equation F_{n + 2}F_n - F_{n + 1}^2 = (-1)^n for each integer n \geq 0.

Basis Step:

Prove P(0). That is:

 F_{0 + 2}F_0 - F_{0 + 1}^2 = (-1)^0 

 F_{2}F_0 - F_{1}^2 = 1 

 (2)(1) - (1)^2 = 1 

 2 - 1 = 1 

 1 = 1 

Therefore P(0) is true.

Inductive Step:

Suppose P(k) for any integer k \geq 0. That is:

 F_{k + 2}F_k - F_{k + 1}^2 = (-1)^k 

This is the inductive hypothesis.

Note that we might need the inductive hypothesis in this form:

 F_{k + 1}^2 = F_{k + 2}F_k - (-1)^k 

Prove P(k + 1), that is:

 F_{(k + 1) + 2}F_{k + 1} - F_{(k + 1) + 1}^2 = (-1)^{k + 1} 

Alternatively:

 F_{k + 3}F_{k + 1} - F_{k + 2}^2 = (-1)^{k + 1} 

Let's evaluate the left-hand side of this equality:

 F_{k + 3}F_{k + 1} - F_{k + 2}^2 

 = (F_{k + 2} + F_{k + 1})F_{k + 1} - F_{k + 2}^2 

 = F_{k + 2}F_{k + 1} + (F_{k + 1}^2) - F_{k + 2}^2 

By the inductive hypothesis, we can substitute thus:

 = F_{k + 2}F_{k + 1} + (F_{k + 2}F_k - (-1)^k) - F_{k + 2}^2 

 = F_{k + 1}(F_{k + 1} + F_k - F_{k + 2}) - (-1)^k 

 = F_{k + 1}((F_{k + 1} + F_k) - F_{k + 2}) - (-1)^k 

 = F_{k + 1}((F_{k + 2}) - F_{k + 2}) - (-1)^k 

 = F_{k + 1}(F_{k + 2} - F_{k + 2}) - (-1)^k 

 = F_{k + 1}(0) - (-1)^k 

 = -(-1)^k 

 = (-1) \cdot (-1)^k 

 = (-1)^{k + 1} 

Q.E.D.

31. Use strong mathematical induction to prove that F_n < 2^n for every integer n \geq 1.

Omitted.

32. Prove that for each integer n \geq 0, \text{gcd}(F_{n + 1}, F_n) = 1. (The definition of \text{gcd} is given in Section 4.10.)

Omitted.

33. It turns out that the Fibonacci sequence satisfies the following explicit formula: For every integer F_n \geq 0,

 F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^{n + 1} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n + 1}\right] 

Verify that the sequence defined by this formula satisfies the recurrence relation F_k = F_{k - 1} + F_{k - 2} for every integer k \geq 2.

Proof:

Let x = \left(\dfrac{1 + \sqrt{5}}{2}\right) and y = \left(\dfrac{1 - \sqrt{5}}{2}\right).

Note that:

 x^2 = \left(\frac{1 + \sqrt{5}}{2}\right)^2 

 = \frac{(1 + \sqrt{5})(1 + \sqrt{5})}{4} 

 = \frac{1 + 2\sqrt{5} + 5 }{4} 

 = \frac{6 + 2\sqrt{5}}{4} 

Similarly, note that:

 y^2 = \left(\frac{1 - \sqrt{5}}{2}\right)^2 

 y^2 = \frac{(1 - \sqrt{5})(1 - \sqrt{5})}{4} 

 y^2 = \frac{1 - 2\sqrt{5} + 5}{4} 

 y^2 = \frac{6 - 2\sqrt{5}}{4} 

Also notice that:

 x + 1 = \left(\frac{1 + \sqrt{5}}{2}\right) + 1 

 x + 1 = \frac{1 + \sqrt{5}}{2} + \frac{2}{2} 

 x + 1 = \frac{3 + \sqrt{5}}{2} 

 x + 1 = \left(\frac{3 + \sqrt{5}}{2}\right)\left(\frac{2}{2}\right) 

 x + 1 = \frac{6 + 2\sqrt{5}}{4} 

 x + 1 = \frac{6 + 2\sqrt{5}}{4} = x^2 

Similarly:

 y + 1 = \left(\frac{1 - \sqrt{5}}{2}\right) + 1 

 = \left(\frac{1 - \sqrt{5}}{2}\right) + \frac{2}{2} 

 = \frac{3 - \sqrt{5}}{2} 

 = \left(\frac{3 - \sqrt{5}}{2}\right)\left(\frac{2}{2}\right) 

 = \frac{6 - 2\sqrt{5}}{4} = y^2 

Suppose k \in \mathbb{Z} and k \geq 2.

We are trying to prove that:

 \frac{1}{\sqrt{5}}[x^k - y^k] = \frac{1}{\sqrt{5}}(x^{k - 1} - y^{k - 1}) + \frac{1}{\sqrt{5}}(x^{k - 2} - y^{k - 2}) 

Since we know that x^2 = x + 1 and y^2 = y + 1, it follows that:

 x^k - y^k 

 = x^2x^{k - 2} - y^2y^{k - 2} 

 = (x + 1)x^{k - 2} - (y + 1)y^{k - 2} 

 = ((x \cdot x^{k - 2}) + (1 \cdot x^{k - 2})) - ((y \cdot y^{k - 2}) + (1 \cdot y^{k - 2}))  

 = x^{k - 1} + x^{k - 2} - y^{k - 1} - y^{k - 2} 

 = x^{k - 1} - y^{k - 1} + x^{k - 2} - y^{k - 2} 

Q.E.D.

34. (For students who have studied calculus) Find \lim\limits_{n \to \infty}\left(\dfrac{F_{n + 1}}{F_n}\right), assuming that the limit exists.

Omitted.

35. (For students who have studied calculus) Prove that \lim\limits_{n \to \infty}\left(\dfrac{F_{n + 1}}{F_n}\right) exists.

Omitted.

36. (For students who have studied calculus) Define x_0, x_1, x_2, \dots as follows:

 x_k = \sqrt{2 + x_{k - 1}} \quad \text{ for each integer } k \geq 1 

 x_0 = 0 

Find \lim\limits_{n \to \infty}x_n. (Assume that the limit exists.)

Omitted.

37. Compound Interest:

Suppose a certain amount of money is deposited in an account paying 4% annual interest compounded quarterly. For each positive integer n, let R_n = \text{ the amount on deposit at the end of the } $n$^th quarter, assuming no additional deposits or withdrawals, and let R_0 be the initial amount deposited.

a. Find a recurrence relation for R_0, R_1, R_2, \dots. Justify your answer.

Since the account pays 4% annual interest compounded quarterly, the total interest is \left(\frac{0.04}{4}\right) = 0.01 or 1%.

Let k \in \mathbb{Z} such that k \geq 0. The recurrence relation can be expressed as:

 R_k = R_{k - 1}+ 0.01(R_{k - 1}) = 1.01R_{k - 1} 

b. If $R_0 = $5,000$, find the am,ount of money on deposit at the end of one year.

R_0 = 5000 \ R_1 = 1.01(5000) = 5050 \ R_2 = 1.01(5050) = 5100.5 \ R_3 = 1.01(5100.5) \approx 5151.51 \ R_4 = 1.01(5151.51) \approx 5203.03 \

c. Find the APY for the account.

 \frac{5203.03 - 5000}{5000} = 0.040606 \text{ or } 4.0606\% 

38. Compound Interest:

Suppose a certain amount of money is deposited in an account paying 3% annual interest compounded monthly. For each positive integer n, let S_n = \text{ the amount on deposit at the end of the } $n$^th month, and let S_0 be the initial amount deposited.

a. Find a recurrence relation for S_0, S_1, S_2, \dots, assuming no additional deposits or withdrawals during the year. Justify your answer.

Since the account pays 3% annual interest compounded monthly, the total interest is \left(\frac{0.03}{12}\right) = 0.0025 or 0.25%.

Let k \in \mathbb{Z} such that k \geq 0. The recurrence relation can be expressed as:

 S_k = S_{k - 1}+ 0.0025(S_{k - 1}) = 1.0025S_{k - 1} 

b. If $S_0 = $10,000$, find the amount of money on deposit at the end of one year.

S_0 = 10000 \ S_1 = 1.0025(10000) = 10025 \ S_2 = 1.0025(10025) \approx 10050.06 \ S_3 = 1.0025(10050.06) \approx 10075.19 \ S_4 = 1.0025(10075.19) \approx 10100.38 \ S_5 = 1.0025(10100.38) \approx 10125.63 \ S_6 = 1.0025(10125.63) \approx 10150.94 \ S_7 = 1.0025(10150.94) \approx 10176.32 \ S_8 = 1.0025(10176.32) \approx 10201.76 \ S_9 = 1.0025(10201.76) \approx 10227.26 \ S_{10} = 1.0025(10227.26) \approx 10252.83 \ S_{11} = 1.0025(10252.83) \approx 10278.46 \ S_{12} = 1.0025(10278.46) \approx 10304.16 \

c. Find the APY for the account.

 \frac{10304.16 - 10000}{10000} = 0.030416 \text{ or } 3.0416\% 

39. With each step you take when climbing a staircase, you can move up either one stair or two stairs. As a result, you can climb the entire staircase taking one stair at a time, taking two at a time, or taking a combination of one-and two-stair increments. For each integer n \geq 1, if the staircase conssits of n stairs, let c_n be the number of different ways to climb the staircase. Find a recurrence relation for c_1, c_2, c_3, \dots. Justify your answer.

Since c_1 = 1 and c_2 = 2, we know that if one climbs to the end of the staircase and there is one step left, then that is n - 1 stairs climbed. If there are two steps left, then that is n - 2 steps climbed. Therefore the recurrence relation can be expressed as:

 c_n = c_{n - 1} + c_{n - 2} 

40. A set of blocks contains blocks of heights 1, 2, and 4 centimeters. Imagine constructing towers by piling blocks of different heights directly on top of one another. (A tower of height 6 cm could be obtained using six $1$-cm blocks, three $2$-cm blocks one $2$-cm block with one $4$-cm block on top, one $4$-cm block with one $2$-cm block on top, and so forth.) Let t_n be the number of ways to construct a tower of height n cm using blocks from the set. (Assume an unlimited supply of blocks of each size.) Find a recurrence relation for t_1, t_2, t_3, \dots. Justify your answer.

Let's establish some initial conditions:

t_1 = 1 \text{ 1 1cm block} \ t_2 = 2 \text{ 2 1cm blocks or 1 2cm block} \ t_3 = 3 \text{ 3 1cm blocks, 1 1cm block and 1 2cm block, or 1 2cm block and 1 1cm block} \

The recurrence relation for n cm blocks then is:

 t_n = t_{n - 1} + t_{n - 2} + t_{n - 4} 

41. Assume the truth of the distributive law (Appendix A, F3), and use the recursive definition of summation, together with mathematical induction, to prove the generalized distributive law that for every positive integer n, if a_1, a_2, \dots, a_n and c are real numbers, then

 \sum_{i = 1}^{n}{ca_i} = c\left(\sum_{i = 1}^{n}{a_i}\right) 

For reference the distributive law states:

For all real numbers a, b, and $c$A

 a(b + c) = ab + ac \quad \text{ and } \quad (b + c)a = ba + ca 

Proof (by mathematical induction):

Let P(n) be the equality:

 \sum_{i = 1}^{n}{ca_i} = c\left(\sum_{i = 1}^{n}{a_i}\right) 

Basis Step:

Prove P(1), that is:

 \sum_{i = 1}^{1}{ca_i} = c\left(\sum_{i = 1}^{1}{a_i}\right) 

Evaluating the left-hand side:

 \sum_{i = 1}^{1}{ca_i} 

 = ca_1 

Evaluating the right-hand side:

 c\left(\sum_{i = 1}^{1}{a_i}\right) 

 = ca_1 

Therefore, since the left-hand and right-hand sides of the equality hold, P(1) is true.

Inductive Step:

Let k \in \mathbb{Z} such that k \geq 1.

Suppose P(k), that is:

 \sum_{i = 1}^{k}{ca_i} = c\left(\sum_{i = 1}^{k}{a_i}\right) 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \sum_{i = 1}^{k + 1}{ca_i} = c\left(\sum_{i = 1}^{k + 1}{a_i}\right) 

By the recursive definition of summation:

 \sum_{i = 1}^{k + 1}{ca_i} = \left(\sum_{i = 1}^{k}{ca_i}\right) + ca_{k + 1} 

Then by the inductive hypothesis, we can substitute the first term:

 = c\left(\sum_{i = 1}^{k}{a_i}\right) + ca_{k + 1} 

By the distributive law:

 = c\left(\sum_{i = 1}^{k}{a_i} + a_{k + 1}\right) 

And then by the recursive definition of summation again:

 = c\left(\sum_{i = 1}^{k + 1}{a_i}\right) 

Therefore P(k + 1) is true.

Q.E.D.

42. Assume the truth of the commutative and associative laws (Appendix A, F1 and F2), and use the recursive definition of product, together with mathematical induction, to prove that for every positive integer n, if a_1, a_2, \dots, a_n and b_1, b_2, \dots, b_n are real numbers, then

 \prod_{i = 1}^{n}{(a_ib_i)} = \left(\prod_{i = 1}^{n}{a_i}\right)\left(\prod_{i = 1}^{n}{b_i}\right) 

For reference the commutative laws state:

For all real numbers a and b,

 a + b = b + a \quad \text{ and } ab = ba 

And the associative laws state:

For all real numbers a, b, and c,

 (a + b) + c = a + (b + c) \quad \text{ and } (ab)c = a(bc) 

Proof (by mathematical induction):

Let P(n) be the equatility:

 \prod_{i = 1}^{n}{(a_ib_i)} = \left(\prod_{i = 1}^{n}{a_i}\right)\left(\prod_{i = 1}^{n}{b_i}\right) 

where n \in \mathbb{Z}^+.

Basis Step:

Prove P(1). That is:

 \prod_{i = 1}^{1}{(a_ib_i)} = \left(\prod_{i = 1}^{1}{a_i}\right)\left(\prod_{i = 1}^{1}{b_i}\right) 

Evaluating the left-hand side:

 \prod_{i = 1}^{1}{(a_ib_i)} 

By the definition of product:

 = a_1 \cdot b_1 

Evaluating the right-hand side:

 \left(\prod_{i = 1}^{1}{a_i}\right)\left(\prod_{i = 1}^{1}{b_i}\right) 

By the recusive definition of product:

 \prod_{i = 1}^{1}{a_i} = a_1 \quad \text{ and } \prod_{i = 1}^{1}{b_i} = b_1 

Therefore:

 = a_1 \cdot b_1 

Therefore, since both sides of the equality hold, P(1) is true.

Inductive Step:

Let k \in \mathbb{Z}^+.

Suppose P(k), that is:

 \prod_{i = 1}^{k}{(a_ib_i)} = \left(\prod_{i = 1}^{k}{a_i}\right)\left(\prod_{i = 1}^{k}{b_i}\right) 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \prod_{i = 1}^{k + 1}{(a_ib_i)} = \left(\prod_{i = 1}^{k + 1}{a_i}\right)\left(\prod_{i = 1}^{k + 1}{b_i}\right) 

Evaluate the left-hand side:

 \prod_{i = 1}^{k + 1}{(a_ib_i)} 

By the recursive definition of product:

 = \left(\prod_{i = 1}^{k}{(a_ib_i)}\right) \cdot a_{k + 1}b_{k + 1} 

By the inductive hypothesis, the first term can be substituted:

 = \left(\prod_{i = 1}^{k}{a_i}\right)\left(\prod_{i = 1}^{k}{b_i}\right) \cdot a_{k + 1}b_{k + 1} 

By the associative laws:

 = \left(\prod_{i = 1}^{k}{a_i}\right) \cdot a_{k + 1} \cdot \left(\prod_{i = 1}^{k}{b_i}\right) \cdot b_{k + 1} 

By the recursive definition of product again:

 = \left(\prod_{i = 1}^{k + 1}{a_i}\right)\left(\prod_{i = 1}^{k + 1}{b_i}\right) 

Which is the right-hand side of the equality. Therefore P(k + 1) is true.

Q.E.D.

43. Assume the truth of the commutative and associative laws (Appendix A, F1 and F2), and use the recursive definition of product, together with mathematical induction, to prove that for each positive integer n, if a_1, a_2, \dots, a_n and c are real numbers, then

 \prod_{i = 1}^{n}{(ca_i)} = c^n\left(\prod_{i = 1}^{n}{a_i}\right) 

For reference the commutative laws state:

For all real numbers a and b,

 a + b = b + a \quad \text{ and } ab = ba 

And the associative laws state:

For all real numbers a, b, and c,

 (a + b) + c = a + (b + c) \quad \text{ and } (ab)c = a(bc) 

Proof (by mathematical induction):

Let P(n) be the equality:

 \prod_{i = 1}^{n}{(ca_i)} = c^n\left(\prod_{i = 1}^{n}{a_i}\right) 

where n \in \mathbb{Z}^+.

Basis Step:

Prove P(1). That is:

 \prod_{i = 1}^{1}{(ca_i)} = c^1\left(\prod_{i = 1}^{1}{a_i}\right) 

Evaluate the left-hand side:

 \prod_{i = 1}^{1}{(ca_i)} 

By the definition of product:

 = ca_1 

 = c^1a_1 

Evaluate the right-hand side:

 c^1\left(\prod_{i = 1}^{1}{a_i}\right) 

By the definition of product:

 = c^1a_1 

Therefore, since the two sides of the equality hold, P(1) is true.

Inductive Step:

Let k \in \mathbb{Z}^+.

Suppose P(k). That is:

 \prod_{i = 1}^{k}{(ca_i)} = c^k\left(\prod_{i = 1}^{k}{a_i}\right) 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 \prod_{i = 1}^{k + 1}{(ca_i)} = c^{k + 1}\left(\prod_{i = 1}^{k + 1}{a_i}\right) 

Evaluating the left-hand side:

 \prod_{i = 1}^{k + 1}{(ca_i)} 

By the definition of recursive product:

 = \left(\prod_{i = 1}^{k}{(ca_i)}\right) \cdot ca_{k + 1} 

By the inductive hypothesis:

 = c^k\left(\prod_{i = 1}^{k}{a_i}\right) \cdot ca_{k + 1} 

By the commutative laws:

 = ca_{k + 1} \cdot c^k\left(\prod_{i = 1}^{k}{a_i}\right) 

By associative laws:

 = c \cdot c^k \cdot \left(\prod_{i = 1}^{k}{a_i}\right) \cdot a_{k + 1} 

By the laws of exponents and by the recursive definition of product:

 = c^{k + 1}\left(\prod_{i = 1}^{k + 1}{a_i}\right) 

This is the right-hand side of our equality. Therefore, P(k + 1) is true.

Q.E.D.

44. The triangle inequality for absolute value states that for all real numbers a and b, |a + b| \leq |a| + |b|. Use the recursive definition of summation, the triangle inequality, the definition of absolute value, and mathematical induction to prove that for each p ositive integer n, if a_1, a_2, \dots, a_n are real numbers, then

 \left| \sum_{i = 1}^{n}{a_i} \right| \leq \sum_{i = 1}^{n}{|a_i|} 

Omitted.

45. Prove that any sum of even integers is even.

Omitted.

46. Prove that any sum of an odd number of odd integers is odd.

Omitted.

47. Deduce from exercise 46 that for any positive integer n if there is a sum of n odd integers that is even, then n is even.

Omitted.

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Page 373

Exercise Set 5.7

1. The formula

 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} 

is true for every integer n \geq 1. Use this fact to solve each of the following problems:

a. If k is an integer and k \geq 2, find a formula for the expression 1 + 2 + 3 + \dots + (k - 1).

n = k - 1

 1 + 2 + 3 + \dots + (k - 1) = \frac{(k - 1)((k - 1) + 1)}{2} 

 = \frac{(k - 1)(k)}{2} 

b. If n is an integer and n \geq 1, find a formula for the expression 5 + 2 + 4 + 6 + 8 + \dots + 2n.

 5 + 2 + 4 + 6 + 8 + \dots + 2n = 5 + 2\left(\frac{(n)(n + 1)}{2}\right) 

 = 5 + n^2 + n 

 = n^2 + n + 5 

c. If n is an integer and n \geq 1, find a formula for the expression 3 + 3 \cdot 2 + 3 \cdot 3 + \dots + 3 \cdot n + n.

 3 + 3 \cdot 2 + 3 \cdot 3 + \dots + 3 \cdot n + n = 3(1 + 2 + 3 + \dots + n) + n 

 = 3\left(\frac{n(n + 1)}{2}\right) + n 

2. The formula

 1 + r + r^2 + \dots + r^n = \frac{r^{n + 1} - 1}{r - 1} 

is true for every real number r except r = 1 and for every integer n \geq 0. Use this fact to solve each of the following problems:

a. If i is an integer and i \geq 1, find a formula for the expression 1 + 2 + 2^2 + \dots + 2^{i - 1}.

 1 + 2 + 2^2 + \dots + 2^{i - 1} = \frac{2^{i - 1 + 1} - 1}{2 - 1} 

 = \frac{2^i - 1}{1} 

 = 2^i - 1 

b. If n is an integer and n \geq 1, find a formula for the expression 3^{n - 1} + 3^{n - 2} + \dots + 3^2 + 3 + 1.

 3^{n - 1} + 3^{n - 2} + \dots + 3^2 + 3 + 1 = \frac{3^{n - 1 + 1} - 1}{3 - 1} 

 = \frac{3^n - 1}{2} 

c. If n is an integer and n \geq 2, find a formula for the expression 2^n + 2^{n - 2} \cdot 3 + 2^{n - 3} \cdot 3 + \dots + 2^2 \cdot 3 + 2 \cdot 3 + 3.

 3 + 3 \cdot 2 + 3 \cdot 2^2 + \dots + 3 \cdot 2^{n - 3} + 3 \cdot 2^{n - 2} + 2^n 

 = 3(2^0 + 2^1 + 2^2 + \dots + 2^{n - 3} + 2^{n - 2}) + 2^n 

 = 2^n + 3\left(\frac{2^{(n - 2) + 1} - 1}{2 - 1}\right) 

 = 2^n + 3\left(\frac{2^{n - 1} - 1}{1}\right) 

 = (2^n) + 3(2^{n - 1} - 1) 

 = (2 \cdot 2^{n - 1}) + 3(2^{n - 1} - 1) 

 = 2 \cdot 2^{n - 1} + 3 \cdot 2^{n - 1} - 3 

 = 5 \cdot 2^{n - 1} - 3 

d. If n is an integer and n \geq 1, find a formula for the expression

Omitted.

In each of 3-15 a sequence is defined recursively. Use iteration to guess an explicit formula for the sequence. Use formulas from Section 5.2 to simplify your answers whenever possible.

3. a_k = ka_{k - 1}, for each integer k \geq 1 a_0 = 1.

 a_0 = 1 

 a_1 = 1 \cdot a_0 = 1 \cdot 1 = 1 

 a_2 = 2 \cdot a_1 = 2 \cdot 1 

 a_3 = 3 \cdot a_2 = 3 \cdot (2 \cdot 1) = 3 \cdot 2 \cdot 1 

 a_4 = 4 \cdot a_3 = 4 \cdot (3 \cdot 2 \cdot 1) = 4 \cdot 3 \cdot 2 \cdot 1 

Guess:

 a_n = n! 

4. b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}}, for each integer k \geq 1 b_0 = 1.

 b_0 = 1 

 b_1 = \frac{b_0}{1 + b_0} = \frac{(1)}{1 + (1)} = \frac{1}{1 + 1} = \frac{1}{2} 

 b_2 = \frac{b_1}{1 + b_1} = \frac{\dfrac{1}{2}}{1 + \left(\dfrac{1}{2}\right)} = \frac{1}{3} 

 b_3 = \frac{b_2}{1 + b_2} = \frac{\dfrac{1}{3}}{1 + \left(\dfrac{1}{3}\right)} = \frac{1}{4} 

 b_4 = \frac{b_3}{1 + b_3} = \frac{\dfrac{1}{4}}{1 + \left(\dfrac{1}{4}\right)} = \frac{1}{5} 

Guess:

 b_n = \frac{1}{n + 1} 

5. c_k = 3c_{k - 1} + 1, for each integer k \geq 2 c_1 = 1.

 c_1 = 1 

 c_2 = 3c_1 + 1 = 3(1) + 1 = 3 + 1 

 c_3 = 3c_2 + 1 = 3(3 + 1) + 1 = (3^2 + 3) + 1 

 c_4 = 3c_3 + 1 = 3(((3^2 + 3) + 1) + 1) + 1 = (3^3 + 3^2 + 3) + 1 

Guess:

 c_n = 3^{n - 1} + 3^{n - 2} + 3^{n - 3} + \dots + 3^3 + 3^2 + 3 + 1 

This is a geometric sequence (Theorem 5.2.2).

 = \frac{3^{(n - 1) + 1} - 1}{3 - 1} 

 = \frac{3^n - 1}{2} 

6. d_k =2d_{k - 1} + 3, for each integer k \geq 2, d_1 = 2.

 d_1 = 2 

 d_2 = 2d_1 + 3 = 2(2) + 3 = 2^2 + 3 

 d_3 = 2d_2 + 3 = 2(2^2 + 3) + 3 = 2^3 + 2 \cdot 3 + 3 

 d_4 = 2d_3 + 3 = 2(2^3 + 2 \cdot 3 + 3) + 3 = 2^4 + 2^2 \cdot 3 + 2 \cdot 3 + 3 

 d_5 = 2d_4 + 3 = 2(2^4 + 2^2 \cdot 3 + 2 \cdot 3 + 3) + 3 = 2^5 + 2^3 \cdot 3 + 2^2 \cdot 3 + 2 \cdot 3 + 3 

 d_5 = 2^5 + 3(2^3 + 2^2 + 2^1 + 2^0) 

 d_5 = 2^5 + 3\sum_{i = 0}^{3}{2^i} 

This is a geometric sequence (Theorem 5.2.2).

Guess:

 d_n = 2^n + 3\sum_{i = 0}^{n - 2}{2^i} 

 d_n = 2^n + 3\frac{2^{(n - 2) + 1} - 1}{2 - 1} 

 = 2^n + 3\frac{2^{n - 1} - 1}{1} 

 = 2^n + 3(2^{n - 1} - 1) 

 = 2^n + 3(2^{n - 1} - 1) 

 = 2^n + 3 \cdot 2^{n - 1} - 3 

 = 2 \cdot 2^{n - 1} + 3 \cdot 2^{n - 1} - 3 

 = 5 \cdot 2^{n - 1} - 3 

7. e_k = 4e_{k - 1} + 5, for each integer k \geq 1 e_0 = 2.

 e_0 = 2 

 e_1 = 4e_0 + 5 = 4 \cdot 2 + 5 

 e_2 = 4e_1 + 5 = 4(4 \cdot 2 + 5) + 5 = 4^2 \cdot 2 + 4 \cdot 5 + 5 

 e_3 = 4e_2 + 5 = 4(4^2 \cdot 2 + 4 \cdot 5 + 5) + 5 = 4^3 \cdot 2 + 4^2 \cdot 5 + 4 \cdot 5 + 5 

 e_4 = 4e_3 + 5 = 4(4^3 \cdot 2 + 4^2 \cdot 5 + 4 \cdot 5 + 5) + 5 = 4^4 \cdot 2 + 4^3 \cdot 5 + 4^2 \cdot 5 + 4 \cdot 5 + 5 

Guess:

 e_n = 4^n \cdot 2 + 4^{n - 1} \cdot 5 + 4^{n - 2} \cdot 5 + \dots + 4 \cdot 5 + 5 

 = 4^n \cdot 2 + 5(4^{n - 1} + 4^{n - 2} + \dots + 4 + 1) 

 = 4^n \cdot 2 + 5\sum_{i = 0}^{n - 1}{4^i} 

 = 4^n \cdot 2 + 5\left(\frac{4^{(n - 1) + 1} - 1}{4 - 1}\right) 

 = 4^n \cdot 2 + 5\left(\frac{4^n - 1}{3}\right) 

 = \frac{3(4^n \cdot 2)}{3} + \left(\frac{5(4^n - 1)}{3}\right) 

 = \frac{3(4^n \cdot 2) + 5(4^n - 1)}{3} 

 = \frac{(6 \cdot 4^n) + (5 \cdot 4^n - 5)}{3} 

 = \frac{6 \cdot 4^n + 5 \cdot 4^n - 5}{3} 

 = \frac{11 \cdot 4^n - 5}{3} 

8. f_k = f_{k - 1} + 2^k, for each integer k \geq 2 f_1 = 1.

 f_1 = 1 

 f_2 = f_1 + 2^2 = (1) + 2^2 = 1 + 2^2 

 f_3 = f_2 + 2^3 = (1 + 2^2) + 2^3 = 1 + 2^2 + 2^3 

 f_4 = f_3 + 2^4 = (1 + 2^2 + 2^3) + 2^4 = 1 + 2^2 + 2^3 + 2^4 

Guess:

 f_n = 1 + \sum_{i = 2}^{n}{2^i} 

 = 1 + \left(\sum_{i = 0}^{n}{2^i} - \sum_{i = 0}^{1}{2^i}\right) 

 = 1 + \frac{2^{n + 1} - 1}{2 - 1} - (2^0 + 2^1) 

 = 1 + 2^{n + 1} - 1 - (1 + 2) 

 = 2^{n + 1} - 3 

9. g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2}, for each integer k \geq 2 g_1 = 1.

 g_1 = 1 

 g_2 = \frac{g_1}{g_1 + 2} = \frac{1}{1 + 2} = \frac{1}{3} = \frac{1}{2^2 - 1} 

 g_3 = \frac{g_2}{g_2 + 2} = \frac{\dfrac{1}{3}}{\dfrac{1}{3} + 2} = \frac{1}{7} = \frac{1}{2^3 - 1} 

 g_4 = \frac{g_3}{g_3 + 2} = \frac{\dfrac{1}{7}}{\dfrac{1}{7} + 2} = \frac{1}{15} = \frac{1}{2^4 - 1} 

Guess:

 g_n = \frac{1}{2^n - 1} 

10. h_k = 2^k - h_{k - 1}, for each integer k \geq 1 h_0 = 1.

 h_0 = 1 

 h_1 = 2^1 - h_0 = 2 - 1 = 2^1 - 2^0 

 h_2 = 2^2 - h_1 = 2^2 - (2^1 - 1) = 2^2 - 2^1 + 2^0 

 h_3 = 2^3 - h_2 = 2^3 - (2^2 - 2^2 + 2^0) = 2^3 - 2^2 + 2^1 - 2^0 

 h_4 = 2^4 - h_3 = 2^4 - (2^3 - 2^2 + 2^1 - 2^0) = 2^4 - 2^3 + 2^2 - 2^1 + 2^0 

Guess:

 h_n = 2^n - 2^{n - 1} + \dots + (-1)^{n - 2} \cdot 2^2 + (-1)^{n - 1} \cdot 2^1 + (-1)^n \cdot 2^0 

 = (-1)^n[(-1)^n \cdot 2^n + \dots + (-1)^2 \cdot 2^2 + (-1)^1 \cdot 2^1 + (-1)^n \cdot 2^0] 

  = (-1)^n[(-2)^n + (-2)^{n - 1} + \dots + (-2)^2 + (-2)^1 + (-2)^0] 

By the definition of a geometric sequence:

 = (-1)^n\left(\frac{(-2)^{n + 1} - 1}{(-2) - 1}\right) 

 = (-1)^n\left(\frac{(-2)^{n + 1} - 1}{-3}\right) 

 = \frac{(-1)^{n + 1}((-2)^{n + 1} - 1)}{(-1)(-3)} 

 = \frac{2^{n + 1} - (-1)^{n + 1}}{3} 

11. p_k = p_{k - 1} + 2 \cdot 3^k, for each integer k \geq 2 p_1 = 2.

 p_1 = 2 

 p_2 = p_1 + 2 \cdot 3^2 = 2 + 2 \cdot 3^2 

 p_3 = p_2 + 2 \cdot 3^3 = (2 + 2 \cdot 3^2) + 2 \cdot 3^3 =  2 + 2 \cdot 3^2 + 2 \cdot 3^3 

Guess:

 p_n = 2 + 2(3^2 + 3^3 + \dots + 3^n) 

 = 2 + 2(3^0 + 3^1 + 3^2 + 3^3 + \dots + 3^n - 1 - 3^1) 

 = 2 + 2\left(\sum_{i = 0}^{n}{3^i} - 1 - 3\right) 

 = 2 + 2\left(\frac{3^{n + 1} - 1}{3 - 1} - 1 - 3\right) 

 = 2 + 2\left(\frac{3^{n + 1} - 1}{2} - 4\right) 

 = 2 + 3^{n + 1} - 1 - 8 

 = 2 + 3^{n + 1} - 9 

 = 3^{n + 1} - 7 

12. s_k = s_{k - 1} + 2k, for each integer k \geq 1 s_0 = 3.

 s_0 = 3 

 s_1 = s_0 + 2(1) = 3 + 2 = 5 

 s_2 = s_1 + 2(2) = (3 + 2) + 2(2) = 3 + 2 + 4 = 9 

 s_3 = s_2 + 2(3) = (3 + 2 + 4) + 2(3) = 3 + 2 + 4 + 6 = 15 

 s_4 = s_3 + 2(4) = (3 + 2 + 4 + 6) + 2(4) = 3 + 2 + 4 + 6 + 8 = 23  

Guess:

 s_n = 3 + 2(1 + 2 + 3 + 4 + \dots + n) 

By Theorem 5.2.1:

 = 3 + 2\left(\frac{n(n + 1)}{2}\right) 

 = 3 + n(n + 1) 

 = 3 + n^2 + n 

 = n^2 + n + 3 

13. t_k = t_{k - 1} + 3k + 1, for each integer k \geq 1 t_0 = 0.

 t_0 = 0 

 t_1 = t_0 + 3(1) + 1 = 0 + 3 + 1 = 3 + 1 = 3 \cdot 1 + 1 

 t_2 = t_1 + 3(2) + 1 = (3 \cdot 1 + 1) + 3 \cdot 2 + 1 = 3 \cdot 1 + 1 + 3 \cdot 2 + 1 

 t_3 = t_2 + 3(3) + 1 = (3 \cdot 1 + 1 + 3 \cdot 2 + 1) + 3 \cdot 3 + 1 = 3 \cdot 1 + 1 + 3 \cdot 2 + 1 + 3 \cdot 3 + 1 

 t_4 = t_3 + 3(4) + 1 = (3 \cdot 1 + 1 + 3 \cdot 2 + 1 + 3 \cdot 3 + 1) + 3 \cdot 4 + 1 

Guess:

 t_n = 3(1 + 2 + 3 + \dots + n) + n 

 = 3\left(\frac{n(n + 1)}{2}\right) + n 

 = \frac{3(n^2 + n)}{2} + \frac{2n}{2} 

 = \frac{3n^2 + 3n + 2n}{2} 

 = \frac{3n^2 + 5n}{2} 

14. x_k = 3x_{k - 1} + k, for each integer k \geq 2 x_1 = 1.

Omitted.

15. y_k = y_{k - 1} + k^2, for each integer k \geq 2 y_1 = 1.

 y_1 = 1 

 y_2 = y_1 + (2)^2 = 1 + 2^2 

 y_3 = y_2 + (3)^2 = (1 + 2^2) + 3^2 = 1 + 2^2 + 3^2 

 y_4 = y_3 + (4)^2 = (1 + 2^2 + 3^2) + 4^2 = 1 + 2^2 + 3^2 + 4^2 

Guess:

 y_n = 1^2 + 2^2 + 3^2 + \dots + n^2 

By Exercise 5.2.10:

 = \frac{n(n + 1)(2n + 1)}{6} 

16. Solve the recurrence relation obtained as the answer to exercise 17(c) of Section 5.6.

The recurrence relation in question is:

 3a_{k - 1} + 2 

For reference:

 a_1 = 2 

Solving:

 a_1 = 2 

 a_2 = 3a_1 + 2 = 3 \cdot 2 + 2 

 a_3 = 3a_2 + 2 = 3 \cdot (3 \cdot 2 + 2) + 2 = 3^2 \cdot 2 + 3 \cdot 2 + 2 

 a_4 = 3a_3 + 2 = 3 \cdot (3^2 \cdot 2 + 3 \cdot 2 + 2) + 2 = 3^3 \cdot 2 + 3^2 \cdot 2 + 3 \cdot 2 + 2 

Guess:

 a_n = 2(3^n + 3^{n - 1} + 3^{n - 2} + \dots + 3^1 + 3^0) 

By the definition of a geometric sequence:

 = 2\left(\frac{3^n - 1}{3 - 1}\right) 

 = 2\left(\frac{3^n - 1}{2}\right) 

 = 3^n - 1 

17. Solve the recurrence relation obtained as the answer to exercise 21(c) of Section 5.6.

The recurrence relation in question is:

 t_n = 3t_{n - 1} + 2 \quad n \geq 2 

For reference:

 t_1 = 2 

 t_2 = 3t_1 + 2 = 3 \cdot 2 + 2 

 t_3 = 3t_2 + 2 = 3 \cdot (3 \cdot 2 + 2) + 2 = 3^2 \cdot 2 + 3 \cdot 2 + 2 

 t_4 = 3t_3 + 2 = 3 \cdot (3^2 \cdot 2 + 3 \cdot 2 + 2) + 2 = 3^3 \cdot 2 + 3^2 \cdot 2 + 3 \cdot 2 + 2 

Guess:

 t_n = 2(3^{n - 1} + 3^{n - 2} + \dots + 3^1 + 3^0) 

By the definition of a geometric sequence (Theorem 5.2.2):

 = 2\left(\frac{3^{(n - 1) + 1} - 1}{3 - 1}\right) 

 = 2\left(\frac{3^n - 1}{2}\right) 

 = 3^n - 1 

18. Suppose d is a fixed constant and a_0, a_1, a_2, \dots is a sequence that satisfies the recurrence relation a_k = a_{k - 1} + d, for each integer k \geq 1. Use mathematical induction to prove that a_n = a_0 + nd, for every integer n \geq 0.

Proof (by mathematical induction):

Let d be any fixed constant, and let a_0, a_1, a_2, \dots be the sequence defined recursively by a_k = a_{k - 1} + d for each integer k \geq 1.

Let P(n) be the equation:

 a_n = a_0 + nd 

We must show by mathematical induction that P(n) is true for every integer n \geq 0.

Basis Step:

Prove that P(0) is true. That is:

 a_0 = a_0 + (0)d 

 a_0 = a_0 

This equality holds, and therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 0.

Suppose P(k). That is:

 a_k = a_0 + kd 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 a_{k + 1} = a_0 + (k + 1)d 

By the definition of the given sequence:

 a_{k + 1} = a_k + d 

By substitution of the inductive hypothesis:

 a_{k + 1} = (a_0 + kd) + d 

By algebra:

 a_{k + 1} = a_0 + (k + 1)d 

Which is the equality that was to be shown. Therefore P(k + 1) is true.

Q.E.D.

19. A worker is promised a bonus if he can increase his productivity by 2 units a day for a period of 30 days. If on day 0 he produces 170 units, how many units must he produce on day 30 to qualify for the bonus?

Let U_n be the number of units produced on day n. Then:

 U_k = U_{k - 1} + 2 

for every integer k \geq 1, and:

 U_0 = 170 

Hence U_0, U_1, U_2, \dots is an arithmetic sequence with a fixed constant 2. It then follows that when n = 30:

 U_n = U_0 + n \cdot 2 

 U_{30} = 170 + (30) \cdot 2 = 170 + 60 = 230 \text{ units} 

Thus, in order to qualify for the bonus, the worker must produce 230 units on day 30.

20. A runner targets herself to improve her time on a certain course by 3 seconds a day. If on day 0 she runs the course in 3 minutes, how fast must she run it on day 14 to stay on target?

First, let's convert 3 minutes to seconds for ease of evaluation:

 3 \text{ minutes } \cdot 60 \frac{\text{seconds}}{\text{minute}} = 180 \text{ seconds} 

Let R_n be the number of seconds the runner ran on day n. Then:

 R_k = R_{k - 1} - 3 

for every integer k \geq 1, and:

 R_0 = 180 

Hence R_0, R_1, R_2, \dots is an arithmetic sequence with a fixed constant 3. It follows then that when n = 14:

 U_n = U_0 - n \cdot 3 

 U_{14} = (180) - 14 \cdot 3 = 180 - 42 = 138 \text{ seconds} 

Therefore, the runner must run the certain course in 138 seconds (approximately 2.3 minutes) on day 14 in order to stay on target.

21. Suppose r is a fixed constant and a_0, a_1, a_2, \dots is a sequence that satisfies the recurrence relation a_k = ra_{k - 1}, for each integer k \geq 1 and a_0 = a. Use mathematical induction to prove that a_n = ar^n, for every integer n \geq 0.

Proof (by mathematical induction):

Let P(n) be the equation:

 a_n = ar^n 

Basis Step:

Prove P(0), that is:

 a_0 = ar^0 

 a_0 = a(1) 

 a_0 = a 

The given problem statement tells us that a_0 = a. Since this matches the equality found for P(0), we can conclude therefore that P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 a_k = ar^k 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 a_{k + 1} = ar^{k + 1} 

By the given recurrence relation:

 a_{k + 1} = r \cdot a_k 

By substitution with the inductive hypothesis:

 a_{k + 1} = r \cdot (a \cdot r^k) 

By algebra:

 a_{k + 1} = ar^{k + 1} 

This equality is what was to be shown, therefore P(k + 1) is true.

Q.E.D.

22. As shown in Example 5.6.8, if a bank pays interest at a rate of i compounded m times a year, then the amount of money P_k at the end of k time periods (where one time period = $\dfrac{1}{m}$^th of a year) satisfies the recurrence relation P_k = \left[1 + \left(\dfrac{1}{m}\right)\right]P_{k - 1} with initial condition P_0 = \text{ the initial amount deposited}. Find an explicit formula for P_n.

 P_0 = P_0 

 P_1 = \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^1 \cdot P_0 

 P_2 = \left[1 + \frac{i}{m}\right] \cdot P_1 = \left[1 + \frac{i}{m}\right] \cdot \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^2 + P_0 

Guess:

 P_n = \left[1 + \frac{i}{m}\right]^n \cdot P_0 

23. Suppose the population of a country increases at a steady rate of 3% per year. If the population is 50 million at a certain time, what will it be 25 years later?

Let P_n be the population of the country at year n. Then:

 P_{k + 1} = 1.03 \cdot P_k 

for every integer k \geq 1, and:

 P_0 = 50000000 

The explicit formula then is:

 P_n = (1.03)^n \cdot P_0 

Then:

 P_{25} = (1.03)^{25} \cdot 50000000 

 \approx 104688896 

Therefore, the population of the country 25 years later will be approximately 104,688896.

24. A chain letter works as follows: One person sends a copy of the letter to five friends, each of whom sends a copy to five friends, each of whom sends a copy to five friends, each of whom sends a copy to five friends, and so forth. How many people will have received copies of the letter after the twentieth reception of this process, assuming no person receives more than one copy?

 \sum_{k = 0}^{20}{5^k} = \frac{5^{21} - 1}{4} 

 \approx 1.192092896 \cdot 10^{14} \text{ people} 

25. A certain computer algorithm executes twice as many operations when it is run with an input size k as when it is run with an input size k - 1 (where k is an integer that is greater than 1). When the algorithm is run with an input size 1, it executes seven operations. How many operations does it execute when it is run with an input size of 25?

Let P_k be the number of operations the algorithm when the input size is k, and P_0 = 7. The recurrence relation is:

 P_k = 2P_{k - 1} 

So:

 P_n = 2^n \cdot P_0 

 P_{25} = 2^{25} \cdot 7 

 = 234881024 

So the algorithm executes 234881024 operations when it is run with an input size of 25.

26. A person saving for retirement makes an initial deposit of $1,000 to a bank account earning interest at a rate of 3% per year compounded monthly, and each month she adds an addition $200 to the account.

a. For each nonnegative integer n, let A_n be the amount in the account at the end of n months. Find the recurrence relation relating A_k to A_{k - 1}.

 A_k = \left[1 + \left(\frac{1}{12}\right)\left(\frac{3}{100}\right)\right] \cdot A_{k - 1} + 200 

 = \left[1 + \left(\frac{3}{1200}\right)\right] \cdot A_{k - 1} + 200 

 = \left[1 + \left(\frac{1}{400}\right)\right] \cdot A_{k - 1} + 200 

 = \frac{401}{400} \cdot A_{k - 1} + 200 

 = 1.0025 \cdot A_{k - 1} + 200 

b. Use iteration to find an explicit formula for A_n.

 A_0 = 1000 

 A_1 = 1.0025 \cdot A_0 + 200 = 1.0025 \cdot (1000) + 200 

 A_2 = 1.0025 \cdot A_1 + 200 = 1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200 

 A_3 = 1.0025 \cdot A_3 + 200 = 1.0025 \cdot (1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200) + 200 

Guess:

 A_n = 200 + 200(1.0025) + 200(1.0025)^2 + \cdots + 200(1.0025)^{n - 1} + 1000(1.0025)^n 

Explicit formula:

 A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n 

c. Use mathematical induction to prove the correctness of the formula you obtained in part (b).

Proof (by mathematical induction):

Let P(n) be the equation:

 A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n 

Basis Step:

Prove P(0), that is:

 A_0 = 200 \cdot \left(\frac{1.0025^0 - 1}{1.0025 - 1}\right) + 1000(1.0025)^0 

 A_0 = 200 \cdot \left(\frac{1 - 1}{0.0025}\right) + 1000(1) 

 A_0 = 200 \cdot 0 + 1000 

 A_0 = 0 + 1000 

 A_0 = 1000 

This equality holds as A_0 was established as being equal to 1000 in the given problem statement. Therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 0.

Suppose P(k), that is:

 A_k = 200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 A_{k + 1} = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} 

By the recurrence relation in part (a), we have:

 A_{k + 1} = 1.0025 \cdot A_k + 200 

By substitution with the inductive hypothesis:

 = 1.0025 \cdot (200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k) + 200 

 = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} + 200 

 = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} + \frac{0.5}{0.0025} 

 = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025 + 0.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} 

 = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{0.0025}\right) + 1000(1.0025)^{k + 1} 

Q.E.D.

d. How much will the account be worth at the end of 20 years? At the end of 40 years?

We can just use the explicit formula:

 A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n 

 A_{20} = 200 \cdot \left(\frac{1.0025^{20} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{20} 

 \approx \$5147.65 

 A_{40} = 200 \cdot \left(\frac{1.0025^{40} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{40} 

 \approx \$9507.67  

e. In how many years will the account be worth $10,000?

 A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n 

 10000 = 200 \cdot \left(\frac{1.0025^n - 1}{0.0025}\right) + 1000(1.0025)^n 

 10000 = \left(\frac{200 \cdot (1.0025^n - 1)}{0.0025}\right) + 1000(1.0025)^n 

 10000 = 80000(1.0025^n - 1) + 1000(1.0025)^n 

 10000 = 80000(1.0025^n) - 80000 + 1000(1.0025)^n 

 10000 = 81000(1.0025^n) - 80000 

 90000 = 81000(1.0025^n) 

 \frac{10}{9} = 1.0025^n 

 \ln\left(\frac{10}{9}\right) = \ln(1.0025^n) 

 \ln\left(\frac{10}{9}\right) = n\ln(1.0025) 

 \frac{\ln\left(\frac{10}{9}\right)}{\ln(1.0025)} = n 

 n \approx 42 \text{ months} 

 \frac{42}{12} = 3.5 \text{ years}  

27. A person borrows $3,000 on a bank credit card at a nominal rate of 18% per year, which is actually charged at a rate of 1.5% per month.

a. What is the annual percentage yield (APY) for the card? (See Example 5.6.8 for a definition of APY.)

 \text{APY} = \left(1 + \frac{r}{n}\right)^n - 1 

 = \left(1 + \frac{0.18}{12}\right)^{12} - 1 

 \approx 0.1956181715 

 \approx 19.6\% 

b. Assume that the person does not place any additional charges on the card and pays the bank $150 each month to pay off the loan. Let B_n be the balance owed on the card after n months. Find an explicit formula for B_n.

 B_0 = 3000 

 B_n = 1.015 \cdot B_{n - 1} - 150 \quad n \geq 1 

 B_1 = 1.015 \cdot B_0 - 150 = 1.015 \cdot 3000 - 150 

 B_2 = 1.015 \cdot B_1 - 150 = 1.015 \cdot (1.015 \cdot 3000 - 150) - 150 

 B_3 = 1.015 \cdot B_2 - 150 = 1.015 \cdot (1.015 \cdot (1.015 \cdot 3000 - 150) - 150) - 150 

Guess:

 B_n = 1.015^n \cdot B_0 - 150 \cdot \left(\frac{1.015^n - 1}{1.015 - 1}\right) 

 = 1.015^n \cdot 3000 - 150 \cdot \left(\frac{1.015^n - 1}{0.015}\right) 

 = 1.015^n \cdot 3000 - 10000(1.015^n - 1) 

 = 1.015^n \cdot 3000 - 10000(1.015^n) + 10000 

 = 10000 - 7000(1.015^n) 

c. How long will be required to pay off the debt?

 0 = 10000 - 7000(1.015^n) 

 7000(1.015^n) = 10000 

 1.015^n = \frac{10000}{7000} 

 1.015^n = \frac{10}{7} 

 \ln(1.015^n) = \ln\left(\frac{10}{7}\right) 

 n\ln(1.015) = \ln\left(\frac{10}{7}\right) 

 n = \frac{\ln\left(\frac{10}{7}\right)}{\ln(1.015)} 

 n \approx 24 \text{ months } = 2 \text{ years} 

d. What is the total amount of money the person will have paid for the loan?

 24 \cdot 150 = \$3600 

In 28-42 use mathematical induction to verify the correctness of the formula you obtained in the referenced exercise.

28. Exercise 3

Let a_0, a_1, a_2, \dots be the sequence defined recursively by a_0 = 1 and a_k = ka_{k - 1} for each integer k \geq 1.

Let the property P(n) be the equation a_n = n!.

Proof by mathematical induction:

Prove P(n) for every integer n \geq 0.

Basis Step:

Prove P(0), that is:

 a_0 = 0! 

 a_0 = 1 

Since 0! = 1, and since by definition of the given sequence, a_0 = 1, the equality holds and therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 a_k = k! 

This is the inductive hypothesis.

Prove P(k + 1). That is:

 a_{k + 1} = (k + 1)! 

By the definition of the given sequence:

 a_k = ka_{k - 1} 

Then:

 a_{k + 1} = (k + 1) \cdot a_k 

By substitution of the inductive hypothesis:

 a_{k + 1} = (k + 1) \cdot k! 

By definition of factorial:

 a_{k + 1} = (k + 1)! 

This is what was to be shown, therefore P(k + 1) is true.

Q.E.D.

29. Exercise 4

Let b_0, b_1, b_2, \dots be the sequence defined recursively by b_0 = 1 and b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}} for each integer k \geq 1.

Let the property P(n) be the equation b_n = \dfrac{1}{n + 1}.

Proof by mathematical induction:

Basis Step:

Prove P(0), that is:

 b_0 = \frac{1}{0 + 1} 

 = \frac{1}{1} 

 = 1 

This equality matches the given value of b_0 = 1, therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 b_k = \dfrac{1}{k + 1} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 b_{k + 1} = \dfrac{1}{(k + 1) + 1} 

Alternatively:

 b_{k + 1} = \dfrac{1}{k + 2} 

By the definition of the given sequence:

 b_k = \dfrac{b_{k - 1}}{1 + b_{k - 1}} 

Then:

 b_{k + 1} = \dfrac{b_k}{1 + b_k} 

By substitution of the inductive hypothesis:

 b_{k + 1} = \frac{\dfrac{1}{k + 1}}{1 + \dfrac{1}{k + 1}} 

 b_{k + 1} = \frac{1}{(k + 1)\left(1 + \dfrac{1}{k + 1}\right)} 

 b_{k + 1} = \frac{1}{k + 1 + 1} 

 b_{k + 1} = \frac{1}{k + 2} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

30. Exercise 5

Let c_1, c_2, c_3, \dots be the sequence defined recursively by c_1 = 1 and c_k = 3c_{k - 1} + 1 for each integer k \geq 2.

Let the property P(n) be the equation c_n = \frac{3^n - 1}{2}.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 c_1 = \frac{3^1 - 1}{2} 

 = \frac{3 - 1}{2} 

 = \frac{2}{2} 

 = 1 

This matches the definition of the given sequence with c_1 = 1. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 c_k = \frac{3^k - 1}{2} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 c_{k + 1} = \frac{3^{k + 1} - 1}{2} 

By the given sequence:

 c_k = 3c_{k - 1} + 1 

Then:

 c_{k + 1} = 3c_k + 1 

By substitution of the inductive hypothesis:

 = 3\left(\frac{3^k - 1}{2}\right) + 1 

 = \frac{3(3^k - 1)}{2} + 1 

 = \frac{3^{k + 1} - 3)}{2} + 1 

 = \frac{3^{k + 1} - 3}{2} + \frac{2}{2} 

 = \frac{3^{k + 1} - 3 + 2}{2} 

 = \frac{3^{k + 1} - 1}{2} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

31. Exercise 6

Let d_1, d_2, d_3, \dots be the sequence defined recursively by d_1 = 2 and d_k = 2d_{k - 1} + 3 for each integer k \geq 2.

Let the property P(n) be the equation d_n = 5 \cdot 2^{n - 1} - 3.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 d_1 = 5 \cdot 2^{1 - 1} - 3 

 = 5 \cdot 2^0 - 3 

 = 5 \cdot 1 - 3 

 = 5 - 3 

 = 2 

This equality matches the given value of d_1 = 2, therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 d_k = 5 \cdot 2^{k - 1} - 3 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 d_{k + 1} = 5 \cdot 2^k - 3 

By the definition of the given sequence:

 d_k = 2d_{k - 1} + 3 

Then:

 d_{k + 1} = 2d_k + 3 

By substitution of the inductive hypothesis:

 = 2(5 \cdot 2^{k - 1} - 3) + 3 

 = 10 \cdot 2^{k - 1} - 6 + 3 

 = 5 \cdot 2 \cdot 2^{k - 1} - 3 

 = 5 \cdot 2^k - 3 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

32. Exercise 7

Let e_0, e_1, e_2, \dots be the sequence defined recursively by e_0 = 2 and e_k = 4e_{k - 1} + 5 for each integer k \geq 1.

Let the property P(n) be the equation e_n = \dfrac{11 \cdot 4^n - 5}{3}.

Proof by mathematical induction:

Basis Step:

Prove P(0), that is:

 e_0 = \dfrac{11 \cdot 4^0 - 5}{3} 

 = \dfrac{11 \cdot 1 - 5}{3} 

 = \dfrac{11 - 5}{3} 

 = \dfrac{6}{3} 

 = 2 

This equality matches the given value of e_0 = 2, therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 e_k = \frac{11 \cdot 4^k - 5}{3} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 e_{k + 1} = \frac{11 \cdot 4^{k + 1} - 5}{3} 

By the given sequence:

 e_k = 4e_{k - 1} + 5 

It follows that:

 e_{k + 1} = 4e_k + 5 

By substitution of the inductive hypothesis:

 = 4\left(\frac{11 \cdot 4^k - 5}{3}\right) + 5 

 = \frac{4(11 \cdot 4^k - 5)}{3} + 5 

 = \frac{44 \cdot 4^k - 20}{3} + 5 

 = \frac{11 \cdot 4 \cdot 4^k - 20}{3} + 5 

 = \frac{11 \cdot 4^{k + 1} - 20}{3} + 5 

 = \frac{11 \cdot 4^{k + 1} - 20}{3} + \frac{15}{3} 

 = \frac{11 \cdot 4^{k + 1} - 20 + 15}{3} 

 = \frac{11 \cdot 4^{k + 1} - 5}{3} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

33. Exercise 8

Let f_1, f_2, f_3, \dots be the sequence defined recursively by f_1 = 1 and f_k = f_{k - 1} + 2^k for each integer k \geq 2.

Let the property P(n) be the equation f_n = 2^{n + 1} - 3.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 f_1 = 2^{1 + 1} - 3 

 = 2^2 - 3 

 = 4 - 3 

 = 1 

This equality matches the give value of f_1 = 1, therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 f_k = 2^{k + 1} - 3 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 f_{k + 1} = 2^{k + 2} - 3 

By the given sequence:

 f_k = f_{k - 1} + 2^k 

It follows that:

 f_{k + 1} = f_k + 2^{k + 1} 

By substitution of the inductive hypothesis:

 = (2^{k + 1} - 3) + 2^{k + 1} 

 = 2^{k + 2} - 3 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

34. Exercise 9

Let g_1, g_2, g_3, \dots be the sequence defined recursively by g_1 = 1 and g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2} for each integer k \geq 2.

Let the property P(n) be the equation g_n = \frac{1}{2^n - 1}.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 g_1 = \frac{1}{2^1 - 1} 

 = \frac{1}{2 - 1} 

 = \frac{1}{1} 

 = 1 

This equality matches the given value of g_1 = 1. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 g_k = \frac{1}{2^k - 1} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 g_{k + 1} = \frac{1}{2^{k + 1} - 1} 

By the given sequence:

 g_k = \dfrac{g_{k - 1}}{g_{k - 1} + 2} 

It follows that:

 g_{k + 1} = \dfrac{g_k}{g_k + 2} 

By substitution of the inductive hypothesis:

 = \dfrac{\dfrac{1}{2^k - 1}}{\left(\dfrac{1}{2^k - 1}\right) + 2} 

 = \dfrac{1}{(2^k - 1)\left(\dfrac{1}{2^k - 1} + 2\right)} 

 = \dfrac{1}{1 + 2(2^k - 1)} 

 = \dfrac{1}{1 + 2^{k + 1} - 2} 

 = \dfrac{1}{2^{k + 1} - 1} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

35. Exercise 10

Let h_0, h_1, h_2, \dots be the sequence defined recursively by h_0 = 1 and h_k = 2^k - h_{k - 1} for each integer k \geq 1.

Let the property P(n) be the equation h_n = \dfrac{2^{n + 1} - (-1)^{n + 1}}{3}.

Proof by mathematical induction:

Basis Step:

Prove P(0), that is:

 h_0 = \frac{2^{0 + 1} - (-1)^{0 + 1}}{3} 

 = \frac{2^1 - (-1)^1}{3} 

 = \frac{2 - (-1)}{3} 

 = \frac{2 + 1}{3} 

 = \frac{3}{3} 

 = 1 

This equality matches the given value of h_0 = 1. Therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 h_k = \frac{2^{k + 1} - (-1)^{k + 1}}{3} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 h_{k + 1} = \frac{2^{k + 2} - (-1)^{k + 2}}{3} 

By the given sequence:

 h_k = 2^k - h_{k - 1} 

It follows that:

 h_{k + 1} = 2^{k + 1} - h_k 

By substitution of the inductive hypothesis:

 = 2^{k + 1} - \left(\frac{2^{k + 1} - (-1)^{k + 1}}{3}\right) 

 = 2^{k + 1} - \frac{2^{k + 1} - (-1)^{k + 1}}{3} 

 = \frac{3 \cdot 2^{k + 1}}{3} - \frac{2^{k + 1} - (-1)^{k + 1}}{3} 

 = \frac{3 \cdot 2^{k + 1} - (2^{k + 1} - (-1)^{k + 1})}{3} 

 = \frac{3 \cdot 2^{k + 1} - 2^{k + 1} + (-1)^{k + 1}}{3} 

 = \frac{2 \cdot 2^{k + 1} + (-1)^{k + 1}}{3} 

 = \frac{2^{k + 2} + (-1)^{k + 1}}{3} 

 = \frac{2^{k + 2} + (-1)(-1)^{k + 2}}{3} 

 = \frac{2^{k + 2} - (-1)^{k + 2}}{3} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

36. Exercise 11

Let p_1, p_2, p_3, \dots be the sequence defined recursively by p_1 = 2 and p_k = p_{k - 1} + 2 \cdot 3^k for each integer k \geq 2.

Let the property P(n) be the equation p_n = 3^{n + 1} - 7.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 p_1 = 3^{1 + 1} - 7 

 = 3^2 - 7 

 = 9 - 7 

 = 2 

This equality matches the given value of p_1 = 2. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 p_k = 3^{k + 1} - 7 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 p_{k + 1} = 3^{k + 2} - 7 

By the given sequence:

 p_k = p_{k - 1} + 2 \cdot 3^k 

It follows that:

 p_{k + 1} = p_k + 2 \cdot 3^{k + 1} 

By substitution of the inductive hypothesis:

 = (3^{k + 1} - 7) + 2 \cdot 3^{k + 1} 

 = -7 + 3 \cdot 3^{k + 1} 

 = -7 + 3^{k + 2} 

 = 3^{k + 2} - 7 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

37. Exercise 12

Let s_0, s_1, s_2, \dots be the sequence defined recursively by s_0 = 3 and s_k = s_{k - 1} + 2k for each integer k \geq 1.

Let the property P(n) be the equation s_n = n^2 + n + 3.

Proof by mathematical induction:

Basis Step:

Prove P(0), that is:

 s_0 = 0^2 + 0 + 3 

 = 0 + 0 + 3 

 = 3 

This equality matches the given value of s_0 = 3. Therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 s_k = k^2 + k + 3 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 s_{k + 1} = (k + 1)^2 + (k + 1) + 3 

Alternatively:

 s_{k + 1} = (k + 1)(k + 1) + k + 4 

 s_{k + 1} = k^2 + 2k + 1 + k + 4 

 s_{k + 1} = k^2 + 3k + 5 

By the given sequence:

 s_k = s_{k - 1} + 2k 

It follows that:

 s_{k + 1} = s_k + 2(k + 1) 

 s_{k + 1} = s_k + 2k + 2 

By substitution of the inductive hypothesis:

 = (k^2 + k + 3) + 2k + 2 

 = k^2 + 3k + 5 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

38. Exercise 13

Let t_0, t_1, t_2, \dots be the sequence defined recursively by t_0 = 0 and t_k = t_{k - 1} + 3k + 1 for each integer k \geq 1.

Let the property P(n) be the equation t_n = \frac{3n^2 + 5n}{2}.

Proof by mathematical induction:

Basis Step:

Prove P(0), that is:

 t_0 = \frac{3(0)^2 + 5(0)}{2} 

 = \frac{3(0) + 0}{2} 

 = \frac{0 + 0}{2} 

 = \frac{0}{2} 

 = 0 

This equality matches the given value of t_0 = 0. Therefore P(0) is true.

Inductive Step:

Let k be any integer such that k \geq 1.

Suppose P(k), that is:

 t_k = \frac{3k^2 + 5k}{2} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 t_{k + 1} = \frac{3(k + 1)^2 + 5(k + 1)}{2} 

Alternatively:

 t_{k + 1} = \frac{3(k + 1)(k + 1) + 5k + 5}{2} 

 t_{k + 1} = \frac{3(k^2 + 2k + 1) + 5k + 5}{2} 

 t_{k + 1} = \frac{3k^2 + 6k + 3 + 5k + 5}{2} 

 t_{k + 1} = \frac{3k^2 + 11k + 8}{2} 

By the given sequence:

 t_k = t_{k - 1} + 3k + 1 

It follows that:

 t_{k + 1} = t_k + 3(k + 1) + 1 

 t_{k + 1} = t_k + 3k + 3 + 1 

 t_{k + 1} = t_k + 3k + 4 

By substitution of the inductive hypothesis:

 = \left(\frac{3k^2 + 5k}{2}\right) + 3k + 4 

 = \frac{3k^2 + 5k}{2} + \frac{6k}{2} + \frac{8}{2} 

 = \frac{3k^2 + 5k + 6k + 8}{2} 

 = \frac{3k^2 + 11k + 8}{2} 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

39. Exercise 14

Let x_1, x_2, x_3, \dots be the sequence defined recursively by x_1 = 1 and x_k = 3x_{k - 1} + k for each integer k \geq 2.

Let the property P(n) be the equation x_n = \frac{1}{4}\left[3^{n + 1} - 3 - 2n\right].

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 x_1 = \frac{1}{4}\left[3^{1 + 1} - 3 - 2(1)\right] 

 = \frac{1}{4}\left[3^2 - 3 - 2\right] 

 = \frac{1}{4}\left[9 - 5\right] 

 = \frac{1}{4}(4) 

 = 1 

This equality matches the given value of x_1 = 1. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 x_k = \frac{1}{4}\left[3^{k + 1} - 3 - 2k\right] 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 3 - 2(k + 1)\right] 

Alternatively:

 x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 3 - 2k - 2\right] 

 x_{k + 1} = \frac{1}{4}\left[3^{k + 2} - 2k - 5\right] 

By the given sequence:

 x_k = 3x_{k - 1} + k 

It follows that:

 x_{k + 1} = 3x_k + k + 1 

By substitution of the inductive hypothesis:

 = 3\left[\frac{1}{4}\left(3^{k + 1} - 3 - 2k\right)\right] + k + 1 

 = \frac{1}{4}\left(3(3^{k + 1} - 3 - 2k)\right) + k + 1 

 = \frac{1}{4}(3^{k + 2} - 9 - 6k) + k + 1 

 = \frac{1}{4}(3^{k + 2} - 9 - 6k) + \frac{4k}{4} + \frac{4}{4} 

 = \frac{1}{4}(3^{k + 2} - 9 - 6k + 4k + 4) 

 = \frac{1}{4}(3^{k + 2} - 5 - 2k) 

 = \frac{1}{4}(3^{k + 2} - 2k - 5) 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

40. Exercise 15

Let y_1, y_2, y_3, \dots be the sequence defined recursively by y_1 = 1 and y_k = y_{k - 1} + k^2 for each integer k \geq 2.

Let the property P(n) be the equation y_n = \frac{n(n + 1)(2n + 1)}{6}.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 y_1 = \frac{1(1 + 1)(2(1) + 1)}{6} 

 = \frac{1(2)(2 + 1)}{6} 

 = \frac{(2)(3)}{6} 

 = \frac{6}{6} 

 = 1 

This equality matches the given value of y_1 = 1. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 y_k = \frac{k(k + 1)(2k + 1)}{6} 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 y_{k + 1} = \frac{(k + 1)(k + 2)(2(k + 1) + 1)}{6} 

Alternatively:

 y_{k + 1} = \frac{(k^2 + 3k + 2)(2k + 2 + 1)}{6} 

 y_{k + 1} = \frac{(k^2 + 3k + 2)(2k + 3)}{6} 

 y_{k + 1} = \frac{k^2(2k + 3) + 3k(2k + 3) + 2(2k + 3)}{6} 

 y_{k + 1} = \frac{2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6}{6} 

 y_{k + 1} = \frac{2k^3 + 9k^2 + 13k + 6}{6} 

By the given sequence:

 y_k = y_{k - 1} + k^2 

It follows that:

 y_{k + 1} = y_k + (k + 1)^2 

By substitution of the inductive hypothesis:

 = \left(\frac{k(k + 1)(2k + 1)}{6}\right) + (k + 1)^2 

 = \frac{k(k + 1)(2k + 1)}{6} + \frac{6(k + 1)^2}{6} 

 = \frac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6} 

 = \frac{k(2k^2 + 3k + 1) + 6(k^2 + 2k + 1)}{6} 

 = \frac{2k^3 + 3k^2 + k + 6k^2 + 12k + 6}{6} 

 = \frac{2k^3 + 9k^2 + 13k + 6}{6} 

This is what was to be shown. Therefore P(k + 1) is true.

41. Exercise 16

Let a_1, a_2, a_3, \dots be the sequence defined recursively by a_1 = 2 and a_k = 3a_{k - 1} + 2 for each integer k \geq 2.

Let the property P(n) be the equation a_n = 3^n - 1.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 a_1 = 3^1 - 1 

 = 3 - 1 

 = 2 

This equality matches the given value of a_1 = 2. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 a_k = 3^k - 1 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 a_{k + 1} = 3^{k + 1} - 1 

By the given sequence:

 a_k = 3a_{k - 1} + 2 

It follows that:

 a_{k + 1} = 3a_k + 2 

By substitution of the inductive hypothesis:

 = 3(3^k - 1) + 2 

 = 3^{k + 1} - 3 + 2 

 = 3^{k + 1} - 1 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

42. Exercise 17

Let t_1, t_2, t_3, \dots be the sequence defined recursively by t_1 = 2 and t_k = 3t_{k - 1} + 2 for each integer k \geq 2.

Let the property P(n) be the equation t_n = 3^n - 1.

Proof by mathematical induction:

Basis Step:

Prove P(1), that is:

 t_1 = 3^1 - 1 

 = 3 - 1 

 = 2 

This equality matches the given value of t_1 = 2. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

 t_k = 3^k - 1 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 t_{k + 1} = 3^{k + 1} - 1 

By the given sequence:

 t_k = 3t_{k - 1} + 2 

It follows that:

 t_{k + 1} = 3t_k + 2 

By substitution of the inductive hypothesis:

 = 3(3^k - 1) + 2 

 = 3^{k + 1} - 3 + 2 

 = 3^{k + 1} - 1 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

In each of 43-49 a sequence is defined recursively. (a) Use iteration to guess an explicit formula for the sequence. (b) Use strong mathematical induction to verify that the formula of part (a) is correct.

43. a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1}, for each integer k \geq 1 a_0 = 2.

a.

 a_0 = 2 

 a_1 = \frac{a_0}{2a_0 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} 

 a_2 = \frac{a_1}{2a_1 - 1} = \frac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} = \frac{2}{4 - 3} = 2 

 a_3 = \frac{a_2}{2a_2 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} 

Guess:

a_n = \begin{cases} 2 & \text{if } n \text{ is even} \ \dfrac{2}{3} & n \text{ is odd} \end{cases}

b.

Let a_0, a_1, a_2, \dots be the sequence defined recursively by a_0 = 2 and a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1} for each integer k \geq 1.

Let the property P(n) be the equation:

a_n = \begin{cases} 2 & \text{if } n \text{ is even} \ \dfrac{2}{3} & \text{if } n \text{ is odd} \end{cases}

Proof by strong mathematical induction:

Basis Step:

Prove P(0) and P(1), that is:

a_0 = \begin{cases} 2 & \text{if } 0 \text{ is even} \ \dfrac{2}{3} & \text{if } 0 \text{ is odd} \end{cases}

and:

a_1 = \begin{cases} 2 & \text{if } 1 \text{ is even} \ \dfrac{2}{3} & \text{if } 1 \text{ is odd} \end{cases}

P(0) is true since the piecewise function tells us that a_0 = 2 since 0 is even and this matches the given value of a_0 = 2.

P(1) is true given the evaluation of a_1 in part (a).

Therefore both P(0) and P(1) are true.

Inductive Step:

Let k be any integer such that k \geq 0. Let i be some integer such that 0 \leq i \leq k.

Suppose P(i), that is:

a_i = \begin{cases} 2 & \text{if } i \text{ is even} \ \dfrac{2}{3} & \text{if } i \text{ is odd} \end{cases}

This is the inductive hypothesis.

Prove P(k + 1), that is:

a_{k + 1} = \begin{cases} 2 & \text{if } k + 1 \text{ is even} \ \dfrac{2}{3} & \text{if } k + 1 \text{ is odd} \end{cases}

By the definition of the sequence:

 a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1} 

It follows that:

 a_{k + 1} = \dfrac{a_k}{2a_k - 1} 

By the inductive hypothesis:

a_{k + 1} = \begin{cases} \dfrac{2}{2(2) - 1} & \text{if } k \text{ is even} \ \dfrac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} & \text{if } k \text{ is odd} \end{cases}

= \ \begin{cases} \dfrac{2}{3} & \text{if } k \text{ is even} \ 2 & \text{if } k \text{ is odd} \end{cases}

It follows that:

= \ \begin{cases} \dfrac{2}{3} & \text{if } k + 1 \text{ is odd} \ 2 & \text{if } k + 1 \text{ is even} \end{cases}

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

44. b_k = \dfrac{2}{b_{k - 1}}, for each integer k \geq 2 b_1 = 1.

a.

 b_1 = 1 

 b_2 = \frac{2}{b_1} = \frac{2}{1} = 2 

 b_3 = \frac{2}{b_2} = \frac{2}{2} = 1 

Guess:

b_n = \begin{cases} 1 & \text{if } n \text{ is odd} \ 2 & \text{if } n \text{ is even} \end{cases}

b.

Let b_1, b_2, b_3, \dots be the sequence defined recursively by b_1 = 1 and b_k = \dfrac{2}{b_{k - 1}} for each integer k \geq 2.

Proof by strong mathematical induction:

Let the property P(n) be the equation:

b_n = \begin{cases} 1 & \text{if } n \text{ is odd} \ 2 & \text{if } n \text{ is even} \end{cases}

Basis Step:

Prove P(1) and P(2).

Both P(1) and P(2) are proven true in part (a).

Inductive Step:

Let k be any integer such that k \geq 2, and let i be some integer such that 1 \leq i \leq k.

Suppose P(i), that is:

b_i = \begin{cases} 1 & \text{if } i \text{ is odd} \ 2 & \text{if } i \text{ is even} \end{cases}

This is the inductive hypothesis.

Prove P(k + 1), that is:

b_{k + 1} = \begin{cases} 1 & \text{if } k + 1 \text{ is odd} \ 2 & \text{if } k + 1 \text{ is even} \end{cases}

By the definition of the given sequence:

 b_k = \dfrac{2}{b_{k - 1}} 

It follows that:

 b_{k + 1} = \dfrac{2}{b_k} 

By the inductive hypothesis:

b_{k + 1} = \begin{cases} \dfrac{2}{1} & \text{if } k \text{ is odd} \ \dfrac{2}{2} & \text{if } k \text{ is even} \end{cases}

= \ \begin{cases} 2 & \text{if } k \text{ is odd} \ 1 & \text{if } k \text{ is even} \end{cases}

= \ \begin{cases} 2 & \text{if } k + 1 \text{ is even} \ 1 & \text{if } k + 1 \text{ is odd} \end{cases}

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

45. v_k = v_{\lfloor \dfrac{k}{2} \rfloor} + v_{\lfloor \dfrac{(k + 1)}{2}\rfloor} + 2, for each integer k \geq 2 v_1 = 1.

a.

 v_1 = 1 

 v_2 = v_{\lfloor \dfrac{2}{2} \rfloor} + v_{\lfloor \dfrac{(2 + 1)}{2} \rfloor} + 2 

 = v_1 + v_1 + 2 = 1 + 1 + 2 

 v_3 = v_{\lfloor \dfrac{3}{2} \rfloor} + v_{\lfloor \dfrac{(3 + 1)}{2} \rfloor} + 2 

 = v_1 + v_2 + 2 = 1 + (1 + 1 + 2) + 2 

 = 3 + 2 \cdot 2 

 v_4 = v_{\lfloor \dfrac{4}{2} \rfloor} + v_{\lfloor \dfrac{(4 + 1)}{2} \rfloor} + 2 

 = v_2 + v_2 + 2 = (1 + 1 + 2) + (1 + 1 + 2) + 2 

 = 4 + 3 \cdot 2 

 v_5 = v_{\lfloor \dfrac{5}{2} \rfloor} + v_{\lfloor \dfrac{(5 + 1)}{2} \rfloor} + 2 

 = v_2 + v_3 + 2 = (1 + 1 + 2) + (3 + 2 \cdot 2) + 2 

 = 5 + 4 \cdot 2 

Guess:

 v_n = n + 2(n - 1) = n + 2n - 2 

 = 3n - 2 

b.

Let v_1, v_2, v_3, \dots be the sequence defined recursively by v_1 = 1 and v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2 for each integer k \geq 2.

Proof by strong mathematical induction:

Let the property P(n) be the equation v_n = 3n - 2.

Basis Step:

Prove P(1), that is:

 v_1 = 3(1) - 2 

 = 3 - 2 

 = 1 

This equality matches the given value of v_1 = 1. Therefore P(1) is true.

Inductive Step:

Let k be any integer such that k \geq 2, and let i be some integer such that 1 \leq i \leq k.

Suppose P(i), that is:

 v_i = 3i - 2 

This is the inductive hypothesis.

Prove P(k + 1), that is:

 v_{k + 1} = 3(k + 1) - 2 

Alternatively:

 v_{k + 1} = 3k + 1 

By the definition of the given sequence:

 v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2 

It follows that:

 v_{k + 1} = v_{\lfloor \frac{(k + 1)}{2} \rfloor} + v_{\lfloor \frac{(k + 2)}{2} \rfloor} + 2 

By the inductive hypothesis:

 = \left(3\lfloor \frac{k + 1}{2} \rfloor - 2\right) + \left(3\lfloor \frac{k + 2}{2}\rfloor - 2\right) + 2 

 = 3\left(\lfloor \frac{k + 1}{2} \rfloor + \lfloor \frac{k + 2}{2} \rfloor \right) - 2 

= \ \begin{cases} 3\left(\frac{k}{2} + \frac{k + 2}{2}\right) - 2 & \text{if } k \text{ is even} \ 3\left(\frac{k + 1}{2} + \frac{k + 1}{2}\right) - 2 & \text{if } k \text{ is odd} \end{cases}

 = 3\left(\frac{2k + 2}{2}\right) - 2 

 = 3(k + 1) - 2 

 = 3k + 3 - 2 

 = 3k + 1 

This is what was to be shown. Therefore P(k + 1) is true.

Q.E.D.

46. s_k = 2s_{k - 2}, for each integer k \geq 2 s_0 = 1, s_1 = 2.

Omitted.

47. t_k = k - t_{k - 1}, for each integer k \geq 1 t_0 = 0.

Omitted.

48. w_k = w_{k - 2} + k, for each integer k \geq 3 w_1 = 1, w_2 = 2.

Omitted.

49. u_k = u_{k - 2} \cdot u_{k - 1}, for each integer k \geq 2 u_0 = u_1 = 2

Omitted.

In 50 and 51 determine whether the given recursively defined sequence satisfies the explicit formula a_n = (n - 1)^2, for every integer n \geq 1.

50. a_k = 2a_{k - 1} + k - 1, for each integer k \geq 2 a_1 = 0.

Omitted.

51. a_k = 4a_{k - 1} - k + 3, for each integer k \geq 2 a_1 = 0.

Omitted.

52. A single line divides a plane into two regions. Two lines (by crossing) can divide a plane into four regions; three lines can divide it into seven regions (see the figure). Let P_n be the maximum number of regions into which n lines divide a plane, where n is a positive integer.

[See Page 375 for image]

a. Derive a recurrence relation for P_k in terms of P_{k - 1}, for each integer k \geq 2.

Omitted.

b. Use iteration to guess an explicit formula for P_n.

Omitted.

53. Compute \left[\begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array}\right]^n for small values of n (up to about 5 or 6). Conjecture explicit formulas for the entries in this matrix, and prove your conjecture using mathematical induction.

Omitted.

54. In economics the behavior of an economy from one period to another is often modeled by recurrence relations. Let Y_k be the income in period k and C_k be the consumption in period k. In one economic model, income in any period is assumed to be the sum of consumption in that period plus investment and government expenditures (which are assumed to be constant from period to period), and consumption in each period is assumed to be a linear function of the income of the preceding period. That is,

 Y_k = C_k + E 

where E is the sum of investment plus government expenditures.

 C_k = c + mY_{k - 1} 

where c and m are constants.

Substituting the second equation into the first gives Y_k = E + c + mY_{k - 1}.

a. Use iteration on the above recurrence relation to obtain

Omitted.

 Y_n = (E + c)\left(\frac{m^n - 1}{m - 1}\right) + m^nY_0 

for every integer n \geq 1.

b. (For students who have studied calculus) Show that if 0 < m < 1, then \lim\limits_{m \to \infty}Y_n = \dfrac{E + c}{1 - m}.

Omitted.

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Page 385

Exercise Set 5.8

1. Which of the following are second-order linear homogeneous recurrence relations with constant coefficients?

a. a_k = 2a_{k - 1} - 5a_{k - 2}

Yes, A = 2 and B = -5

b. b_k = kb_{k - 1} + b_{k - 2}

No, A = k, which is not a constant coefficient.

c. c_k = 3c_{k - 1} \cdot c_{k - 2}^2

No, this does not take the form of a linear homogeneous recurrence relation with constant coefficients. The first term 3c_{k - 1} is fine, but then there is multiplication instead of addition/subtraction, and c_{k - 2}^2 violates the homogeneous rule (each term must have the same total degree).

d. d_k = 3d_{k - 1} + d_{k - 2}

Yes, A = 3, B = 1.

e. r_k = r_{k - 1} - r_{k - 2} - 2

No, this violates the "second-order" definition, as r_k must only contain the two previous terms r_{k - 1} and r_{k - 2}, the -2 at the end of the equation violates this definition.

f. s_k = 10s_{k - 2}

Yes.

2. Which of the following are second-order linear homogeneous recurrence relations with constant coefficients?

a. a_k = (k - 1)a_{k - 1} + 2ka_{k - 2}

No.

b. b_k = -b_{k - 1} + 7b_{k - 2}

Yes.

c. c_k = 3c_{k - 1} + 1

No.

d. d_k = 3d_{k - 1}^2 + d_{k - 2}

No.

e. r_k = r_{k - 1} + 6r_{k - 3}

Yes.

f. s_k = s_{k - 1} + 10s_{k - 2}

Yes.

3. Let a_0, a_1, a_2, \dots be the sequence defined by the explicit formula

 a_n = C \cdot 2^n + D \quad \text{ for every integer } n \geq 0 

where C and D are real numbers.

a. Find C and D so that a_0 = 1 and a_1 = 3. What is a_2 in this case?

 a_0 = 1 = C \cdot 2^0 + D = C(1) + D = C + D 

 a_1 = 3 = C \cdot 2^1 + D = 2C + D 

So we have:

 C + D = 1 \quad \text{ and } 2C + D = 3 

Let's solve:

 C = 1 - D 

 2(1 - D) + D = 3 

 2 - 2D + D = 3 

 2 - D = 3 

 2 = 3 + D 

 -1 = D 

 C = 1 - (-1) 

 C = 2 

So:

 a_n = C \cdot 2^n + D 

 a_n = 2 \cdot 2^n + (-1) 

 a_n = 2^{n + 1} - 1 

And:

 a_2 = 2^{2 + 1} - 1 

 a_2 = 2^{3} - 1 

 a_2 = 8 - 1 

 a_2 = 7 

b. Find C and D so that a_0 = 0 and a_1 = 2. What is a_2 in this case?

 a_0 = 0 = C \cdot 2^0 + D = C + D 

 a_1 = 2 = C \cdot 2^1 + D = 2C + D 

So we have:

 C + D = 0 \quad \text{ and } \quad 2C + D = 2 

Let's solve:

 C = -D 

 2(-D) + D = 2 

 -2D + D = 2 

 -D = 2 

 D = -2 

 C = -(-2) 

 C = 2 

Then:

 a_n = 2 \cdot 2^n + (-2) 

 a_n = 2^{n + 1} - 2 

Then:

 a_2 = 2^{2 + 1} - 2 

 a_2 = 2^3 - 2 

 a_2 = 8 - 2 

 a_2 = 6 

4. Let b_0, b_1, b_2, \dots be the sequence defined by the explicit formula

 b_n = C \cdot 3^n + D(-2)^n \quad \text{ for each integer } n \geq 0 

where C and D are real numbers.

a. Find C and D so that b_0 = 0 and b_1 = 5. What is b_2 in this case?

 b_0 = 0 = C \cdot 3^0 + D(-2)^0 = C + D 

 b_1 = 5 = C \cdot 3^1 + D(-2)^1 = 3C - 2D 

So we have:

 C + D = 0 \quad \text{ and } \quad 3C - 2D = 5 

Let's solve:

 C = -D 

 3(-D) - 2D = 5 

 -3D - 2D = 5 

 -5D = 5 

 D = -1 

 C = -(-1) 

 C = 1 

So we have:

 b_n = 1 \cdot 3^n + (-1)(-2)^n 

 b_n = 3^n - (-2)^n 

Then:

 b_2 = 3^2 - (-2)^2 

 b_2 = 9 - (4) 

 b_2 = 5 

b. Find C and D so that b_0 = 3 and b_1 = 4. What is b_2 in this case?

 b_0 = 3 = C \cdot 3^0 + D(-2)^0 = C + D 

 b_1 = 4 = C \cdot 3^1 + D(-2)^1 = 3C - 2D 

So we have:

 C + D = 3 \quad \text{ and } \quad 3C - 2D = 4 

Let's solve:

 C = 3 - D 

 3(3 - D) - 2D = 4 

 9 - 3D - 2D = 4 

 9 - 5D = 4 

 9 = 4 + 5D 

 5 = 5D 

 1 = D 

 C = 3 - 1 

 C = 2 

Then we have:

 b_n = 2 \cdot 3^n + (1)(-2)^n 

 b_n = 2 \cdot 3^n + (-2)^n 

Then we have:

 b_2 = 2 \cdot 3^2 + (-2)^2 

 b_2 = 2 \cdot 9 + (4) 

 b_2 = 18 + 4 

 b_2 = 22 

5. Let a_0, a_1, a_2, \dots be the sequence defined by the explicit formula

 a_n = C \cdot 2^n + D \quad \text{ for each integer } n \geq 0 

where C and D are real numbers. Show that for any choice of C and D,

 a_k = 3a_{k - 1} - 2a_{k - 2} \quad \text{ for every integer } k \geq 2 

Proof:

Let a_0, a_1, a_2, \dots be the sequence defined by the explicit formula a_n = C \cdot 2^n + D for each integer n \geq 0, where C and D are real numbers.

Let k be any integer such that k \geq 2. It follows that:

 a_k = C \cdot 2^k + D 

 a_{k - 1} = C \cdot 2^{k - 1} + D 

 a_{k - 2} = C \cdot 2^{k - 2} + D 

We must show that for any choice of C and D, that:

 a_k = 3a_{k - 1} - 2a_{k - 2} 

By substitution:

 = 3(C \cdot 2^{k - 1} + D) - 2(C \cdot 2^{k - 2} + D) 

Then, by algebra:

 = 3C \cdot 2^{k - 1} + 3D - 2C \cdot 2^{k - 2} - 2D 

 = 3C \cdot 2^{k - 1} - 2C \cdot 2^{k - 2} + D 

 = 3C \cdot 2^{k - 1} - C \cdot 2 \cdot 2^{k - 2} + D 

 = 3C \cdot 2^{k - 1} - C \cdot 2^{k - 1} + D 

 = 2C \cdot 2^{k - 1} + D 

 = C \cdot 2 \cdot 2^{k - 1} + D 

 = C \cdot 2^k + D 

By the definition of the given equation:

 = a_k 

This is what was to be shown.

Q.E.D.

6. Let b_0, b_1, b_2, \dots be the sequence defined by the explicit formula

 b_n = C \cdot 3^n + D(-2)^n \quad \text{ for every integer } n \geq 0 

where C and D are real numbers. Show that for any choice of C and D,

 b_k = b_{k - 1} + 6b_{k - 2} \quad \text{ for each integer } k \geq 2 

Proof:

Let b_0, b_1, b_2, \dots be the sequence defined by the explicit formula b_n = C \cdot 3^n + D(-2)^n for every integer n \geq 0, where C and D are real numbers.

Let k be any integer where k \geq 2. It then follows that:

 b_k = C \cdot 3^k + D(-2)^k 

 b_{k - 1} = C \cdot 3^{k - 1} + D(-2)^{k - 1} 

 b_{k - 2} = C \cdot 3^{k - 2} + D(-2)^{k - 2} 

We must show that for any choice C and D:

 b_k = b_{k - 1} + 6b_{k - 2} 

By substitution:

 b_{k - 1} + 6b_{k - 2} = (C \cdot 3^{k - 1} + D(-2)^{k - 1}) + 6(C \cdot 3^{k - 2} + D(-2)^{k - 2}) 

By algebra:

 = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 6C \cdot 3^{k - 2} + 6D(-2)^{k - 2} 

 = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3 \cdot 3^{k - 2} + (-3)D \cdot (-2) \cdot (-2)^{k - 2} 

 = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3^{k - 1} + (-3)D \cdot (-2)^{k - 1} 

 = 3C \cdot 3^{k - 1} + (-2)D(-2)^{k - 1} 

 = C \cdot 3 \cdot 3^{k - 1} + D \cdot (-2) \cdot (-2)^{k - 1} 

 = C \cdot 3^k + D(-2)^k 

 = b_k 

This is what was to be shown.

Q.E.D.

7. Solve the system of equations in Example 5.8.4 to obtain

 C = \frac{1 + \sqrt{5}}{2\sqrt{5}} \quad \text{ and } \quad D = \frac{-(1 - \sqrt{5})}{2\sqrt{5}} 

Proof:

The initial conditions are:

 F_0 = F_1 = 1 

5.8.4 has established that the Fibonacci relation is a second-order linear homogeneous recurrence relation with constant coefficients. It also established that the explicit formula for the Fibonacci sequence is:

 F_n = C\left(\frac{1 + \sqrt{5}}{2}\right)^n + D\left(\frac{1 - \sqrt{5}}{2}\right)^n 

It follows then that:

 F_0 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^0 + D\left(\frac{1 - \sqrt{5}}{2}\right)^0 

 F_0 = 1 = C + D 

 F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^1 + D\left(\frac{1 - \sqrt{5}}{2}\right)^1 

 F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) 

So we have:

 C + D = 1 

 C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 

Evaluating for both C and D:

 C = 1 - D 

 (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 

 (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) = 1 - D\left(\frac{1 - \sqrt{5}}{2}\right) 

 \frac{(1 - D)(1 + \sqrt{5})}{2} = \frac{2}{2} - \left(\frac{D(1 - \sqrt{5})}{2}\right) 

 \frac{(1 - D)(1 + \sqrt{5})}{2} = \left(\frac{2 - D(1 - \sqrt{5})}{2}\right) 

 (1 - D)(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) 

 (1 \cdot 1) + (-D)(1) + (1)(\sqrt{5}) + (-D)(\sqrt{5}) = 2 - D(1 - \sqrt{5}) 

 1 - D + \sqrt{5} - D\sqrt{5} = 2 - D(1 - \sqrt{5}) 

 (1 + \sqrt{5}) - D(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) 

 -D(1 + \sqrt{5}) + D(1 - \sqrt{5}) = 2 - (1 + \sqrt{5}) 

 D(-1(1 + \sqrt{5}) + (1 - \sqrt{5})) = 2 - 1 - \sqrt{5} 

 D((1 - \sqrt{5}) - (1 + \sqrt{5})) = 1 - \sqrt{5} 

 D(1 - \sqrt{5} - 1 - \sqrt{5}) = 1 - \sqrt{5} 

 D(-\sqrt{5} - \sqrt{5}) = 1 - \sqrt{5} 

 D(-2\sqrt{5}) = 1 - \sqrt{5} 

 D = \frac{1 - \sqrt{5}}{-2\sqrt{5}} 

 D = \frac{\sqrt{5} - 1}{2\sqrt{5}} 

 D = \frac{5 - \sqrt{5}}{2 \cdot 5} 

 D = \frac{5 - \sqrt{5}}{10} 

Then:

 C = 1 - \left(\frac{5 - \sqrt{5}}{10}\right) 

 C = \frac{10}{10} - \left(\frac{5 - \sqrt{5}}{10}\right) 

 C = \frac{10 - (5 - \sqrt{5})}{10} 

 C = \frac{10 - 5 + \sqrt{5}}{10} 

 C = \frac{5 + \sqrt{5}}{10} 

The closed form solution then is:

 F_n = \left(\frac{5 + \sqrt{5}}{10}\right)\left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{5 - \sqrt{5}}{10}\right)\left(\frac{1 - \sqrt{5}}{2}\right)^n 

In each of 8-10: (a) suppose a sequence of the form 1, t, t^2, t^3, \dots, t^n, \dots where t \neq 0, satisfies the given recurrence relation (but not necessarily the initial conditions), and find all possible values of t: (b) suppose a sequence satisfies the given initial conditions as well as the recurrence relation, and find an explicit formula for the sequence.

8. a_k = 2a_{k - 1} + 3a_{k - 2}, for every integer k \geq 2 a_0 = 1, a_1 = 2

a.

Since the given second-order homogeneous recurrence relation with constant coefficient, a_k, is satisfied by the sequence 1, t, t^2, t^3, \dots t^n, \dots. t^k can be expressed as:

 t^k = 2t^{k - 1} + 3t^{k - 2} 

It follows that:

 t^2 = 2t^{2 - 1} + 3t^{2 - 2} 

 t^2 = 2t^1 + 3t^0 

 t^2 = 2t + 3 

By algebra then:

 t^2 - 2t - 3 = 0 

 (t - 3)(t + 1) = 0 

Therefore the possible values of t are:

 t = -1 \quad \text{ and } \quad t = 3 

b.

It follows from (a) and the distinct roots theorem that the explicit formula for the given sequence follows the form of:

 a_n = Cr^n + Ds^n 

where r and s are the roots found in part (a):

 a_n = C \cdot 3^n + D \cdot (-1)^n 

for every integer n \geq 0.

By the Distinct roots theorem we can find C and D by looking at the values of a_0 and a_1 and evaluating for them:

 a_0 = 1 = C \cdot 3^0 + D \cdot (-1)^0 = C + D 

 a_1 = 2 = C \cdot 3^1 + D \cdot (-1)^1 = 3C - D 

So we have:

 C + D = 1 \quad \text{ and } \quad 3C - D = 2 

Evaluating:

 C = 1 - D 

 3(1 - D) - D = 2 

 3 - 3D - D = 2 

 3 - 4D = 2 

 3 = 2 + 4D 

 1 = 4D 

 \frac{1}{4} = D 

 C = 1 - \frac{1}{4} 

 C = \frac{4}{4} - \frac{1}{4} 

 C = \frac{3}{4} 

So, our explicit formula for the given sequence is:

 a_n = \frac{3}{4} \cdot 3^n + \frac{1}{4} \cdot (-1)^n 

9. b_k = 7b_{k - 1} - 10b_{k - 2}, for every integer k \geq 2 b_0 = 2, b_1, = 2

a.

Since the given second-order homogeneous recurrence relation with constant coefficient, b_k, is satisfied by the sequence 1, t, t^2, t^3, \dots t^n, \dots. t^k can be expressed as:

 t^k = 7t^{k - 1} - 10t^{k - 2} 

It follows that:

 t^2 = 7t^{2 - 1} - 10t^{2 - 2} 

 t^2 = 7t^1 - 10t^0 

 t^2 = 7t - 10 

By algebra:

 t^2 - 7t + 10 = 0 

 (t - 5)(t - 2) = 0 

The possible values of t then are:

 t = 2 \quad \text{ and } \quad t = 5 

b.

It follows from (a) and the distinct roots theorem that the explicit formula for the given sequence follows the form of:

 b_n = Cr^n + Ds^n 

where r and s are the roots found in part (a):

 b_n = C \cdot 2^n + D \cdot 5^n 

for every integer n \geq 0.

By the Distinct roots theorem we can find C and D by looking at the values of b_0 and b_1 and evaluating for them:

 b_0 = 2 = C \cdot 2^0 + D \cdot 5^0 = C + D 

 b_1 = 2 = C \cdot 2^1 + D \cdot 5^1 = 2C + 5D 

So we have:

 C + D = 2 \quad \text{ and } \quad 2C + 5D = 2 

Evaluating:

 C = 2 - D 

 2(2 - D)  + 5D = 2 

 4 - 2D  + 5D = 2 

 4 + 3D = 2 

 3D = 2 - 4 

 3D = -2 

 D = -\frac{2}{3} 

 C = 2 - \left(-\frac{2}{3}\right) 

 C = \frac{6}{3} + \frac{2}{3} 

 C = \frac{8}{3} 

So the explicit formula for the given sequence is:

 b_n = \frac{8}{3} \cdot 2^n + \left(-\frac{2}{3}\right) \cdot 5^n 

10. c_k = c_{k - 1} + 6c_{k - 2}, for every integer k \geq 2 c_0 = 0, c_1 = 3

a.

Since the given second-order homogeneous recurrence relation with constant coefficient, c_k, is satisfied by the sequence 1, t, t^2, t^3, \dots t^n, \dots. t^k can be expressed as:

 t^k = t^{k - 1} + 6t^{k - 2} 

It follows that:

 t^2 = t^{2 - 1} + 6t^{2 - 2} 

 t^2 = t^1 + 6t^0 

 t^2 = t + 6 

By algebra:

 t^2 - t - 6 = 0 

 (t - 3)(t + 2) = 0 

So the possible values of t are:

 t = -2 \quad \text{ and } \quad t = 3 

b.

It follows from (a) and the distinct roots theorem that the explicit formula for the given sequence follows the form of:

 c_n = Cr^n + Ds^n 

where r and s are the roots found in part (a):

 c_n = C \cdot 3^n + D \cdot (-2)^n 

for every integer n \geq 0.

By the Distinct roots theorem we can find C and D by looking at the values of c_0 and c_1 and evaluating for them:

 c_0 = 0 = C \cdot 3^0 + D \cdot (-2)^0 = C + D 

 c_1 = 3 = C \cdot 3^1 + D \cdot (-2)^1 = 3C - 2D 

So we have:

 C + D = 0 \quad \text{ and } \quad 3C - 2D = 3 

Evaluating:

 C = -D 

 3(-D) - 2D = 3 

 -3D - 2D = 3 

 -5D = 3 

 D = -\frac{3}{5} 

 C = -\left(-\frac{3}{5}\right) 

 C = \frac{3}{5} 

So our explicit formula for the given sequence is:

 c_n = \frac{3}{5} \cdot 3^n + \left(-\frac{3}{5}\right) \cdot (-2)^n 

In each of 11-16 suppose a sequence satisfies the given recurrence relation and initial conditions. Find an explicit formula for the sequence.

11. d_k = 4d_{k - 2} , for each integer k \geq 2 d_0 = 1, d_1 = -1

A lot of this is similar to 8-10. Let's streamline it:

 t^k = 4t^{k - 2} 

 t^2 = 4t^{2 - 2} 

 t^2 = 4t^0 

 t^2 = 4 

 t = \pm 2 

 d_n = C \cdot r^n + D \cdot s^n 

 d_n = C \cdot 2^n + D \cdot (-2)^n 

 d_0 = 1 = C \cdot 2^0 + D \cdot (-2)^0 = C + D 

 d_1 = -1 = C \cdot 2^1 + D \cdot (-2)^1 = 2C - 2D 

 C + D = 1 \quad \text{ and } \quad 2C - 2D = -1 

 C = 1 - D 

 2(1 - D) - 2D = -1 

 2 - 2D - 2D = -1 

 2 - 4D = -1 

 2 = -1 + 4D 

 3 = 4D 

 \frac{3}{4} = D 

 C = 1 - \frac{3}{4} 

 C = \frac{4}{4} - \frac{3}{4} 

 C = \frac{1}{4} 

 \boxed{d_n = \frac{1}{4} \cdot 2^n + \frac{3}{4} \cdot (-2)^n} 

12. e_k = 9e_{k - 1}, for each integer k \geq 2 e_0 = 0, e_1 = 2

 t^k = 9t^{k - 1} 

 t^2 = 9t^{2 - 1} 

 t^2 = 9t^1 

 t^2 = 9t 

 t^2 - 9t = 0 

 t(t - 9) = 0 

This is a single root: t = 9. And so takes the form:

 e_n = Cr^n + Dnr^n 

 e_n = C \cdot 9^n + Dn \cdot 9^n 

 e_0 = 0 = C \cdot 9^0 + D(0) \cdot 9^0 = C 

 e_1 = 2 = C \cdot 9^1 + D(1) \cdot 9^1 = 9C + 9D 

 C = 0 \quad \text{ and } 9C + 9D = 2 

 9(0) + 9D = 2 

 0 + 9D = 2 

 9D = 2 

 D = \frac{2}{9} 

 e_n = 0 \cdot 9^n + \left(\frac{2}{9}\right)n \cdot 9^n 

 \boxed{e_n = \frac{2}{9}n \cdot 9^n} 

13. r_k = 2r^{k - 1} - r^{k - 2}, for each integer k \geq 2 r_0 = 1, r_1 = 4

 t^k = 2t^{k - 1} - t^{k - 2} 

 t^2 = 2t^{2 - 1} - t^{2 - 2} 

 t^2 = 2t^1 - t^0 

 t^2 = 2t - 1 

 t^2 - 2t + 1 = 0 

 (t - 1)(t - 1) = 0 

 t = 1 

 r_n = Cs^n + Dns^n 

 r_n = C \cdot 1^n + D(n) \cdot 1^n 

 r_0 = 1 = C \cdot 1^0 + D(0) \cdot 1^0 = C 

 r_1 = 4 = C \cdot 1^1 + D(1) \cdot 1^1 = C + D 

 C = 1 \quad \text{ and } \quad C + D = 4 

 1 + D = 4 

 D = 3 

 r_n = 1 \cdot 1^n + 3(n) \cdot 1^n 

 r_n = 1^n + 3n \cdot 1^n 

 \boxed{r_n = 1 + 3n} 

14. s_k = -4s_{k - 1} - 4s_{k - 2}, for every integer k \geq 2 s_0 = 0, s_1 = -1

 t^k = -4t^{k - 1} - 4t^{k - 2} 

 t^2 = -4t^{2 - 1} - 4t^{2 - 2} 

 t^2 = -4t^1 - 4t^0 

 t^2 = -4t - 4 

 t^2 + 4t + 4 = 0 

 (t + 2)(t + 2) = 0 

 t = -2 

 s_n = Cr^n + Dnr^n 

 s_n = C \cdot (-2)^n + Dn \cdot (-2)^n 

 s_0 = 0 = C \cdot (-2)^0 + D(0) \cdot (-2)^0 = C 

 s_1 = -1 = C \cdot (-2)^1 + D(1) \cdot (-2)^1 = -2C - 2D 

 C = 0 \quad \text{ and } \quad -2C - 2D = -1 

 -2(0) - 2D = -1 

 -2D = -1 

 D = \frac{1}{2} 

 s_n = 0 \cdot (-2)^n + \frac{1}{2}n \cdot (-2)^n 

 \boxed{s_n = \frac{1}{2}n \cdot (-2)^n} 

15. t_k = 6t_{k - 1} - 9t_{k - 2}, for each integer k \geq 2 t_0 = 1, t_1= 3

 x^k = 6x^{k - 1} - 9x^{k - 2} 

 x^2 = 6x^{2 - 1} - 9x^{2 - 2} 

 x^2 = 6x^1 - 9x^0 

 x^2 = 6x - 9 

 x^2 - 6x + 9 = 0 

 (x - 3)(x - 3) = 0 

 x = 3 

 t_n = Cr^n + Dnr^n 

 t_n = C \cdot 3^n + Dn \cdot 3^n 

 t_0 = 1 = C \cdot 3^0 + D(0) \cdot 3^0 = C 

 t_1 = 3 = C \cdot 3^1 + D(1) \cdot 3^1 = 3C + 3D 

 C = 1 \quad \text{ and } \quad 3C + 3D = 3 

 3(1) + 3D = 3 

 3 + 3D = 3 

 3D = 0 

 D = 0 

 t_n = 1 \cdot 3^n + (0)n \cdot 3^n 

 \boxed{t_n = 3^n} 

16. s_k = 2s_{k - 1} + 2s_{k - 2}, for every integer k \geq 2 s_0 = 1, s_1 = 3

 t^k = 2t^{k - 1} + 2t^{k - 2} 

 t^2 = 2t^{2 - 1} + 2t^{2 - 2} 

 t^2 = 2t^1 + 2t^0 

 t^2 = 2t + 2 

 t^2 - 2t - 2 = 0 

 t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} 

 t = \frac{2 \pm \sqrt{12}}{2} 

 t = \frac{2 \pm 2\sqrt{3}}{2} 

 t = 1 \pm \sqrt{3} 

 s_n = Cx^n + Dy^n 

 s_n = C \cdot \left(1 + \sqrt{3}\right)^n + D \cdot \left(1 - \sqrt{3}\right)^n 

 s_0 = 1 = C \cdot \left(1 + \sqrt{3}\right)^0 + D \cdot \left(1 - \sqrt{3}\right)^0 = C + D 

 s_1 = 3 = C \cdot \left(1 + \sqrt{3}\right)^1 + D \cdot \left(1 - \sqrt{3}\right)^1 = C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right)  

 C + D = 1 \quad \text{ and } \quad C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 

 C = 1 - D 

 (1 - D) \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 

 (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 (1)(1) + (-D)(1) + (1)(\sqrt{3}) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 1 + (-D)(1) + \sqrt{3} + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 (1 + \sqrt{3}) + (-D)(1) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 (1 + \sqrt{3}) + (-D)(1 + \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 (1 + \sqrt{3}) + D(-1 - \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] 

 D(-1 - \sqrt{3}) + D \cdot \left(1 - \sqrt{3}\right) = 3 - (1 - \sqrt{3}) 

 D\left[(-1 - \sqrt{3}) + \left(1 - \sqrt{3}\right)\right] = 3 - 1 + \sqrt{3} 

 D\left[-1 - \sqrt{3} + 1 - \sqrt{3}\right] = 2 + \sqrt{3} 

 D\left[-\sqrt{3} - \sqrt{3}\right] = 2 + \sqrt{3} 

 D(-2\sqrt{3}) = 2 + \sqrt{3} 

 D = \frac{2 + \sqrt{3}}{-2\sqrt{3}} 

 D = \frac{2\sqrt{3} + 3}{-2 \cdot 3} 

 D = \frac{2\sqrt{3} + 3}{-6} 

 C = 1 - \left(\frac{2\sqrt{3} + 3}{-6}\right) 

 C = \frac{-6}{-6} - \left(\frac{2\sqrt{3} + 3}{-6}\right) 

 C = \frac{-6 - (2\sqrt{3} + 3)}{-6} 

 C = \frac{-6 - 2\sqrt{3} - 3}{-6} 

 C = \frac{-9 - 2\sqrt{3}}{-6} 

 C = \frac{9 + 2\sqrt{3}}{6} 

 s_n = \frac{9 + 2\sqrt{3}}{6} \cdot \left(1 + \sqrt{3}\right)^n + \frac{2\sqrt{3} + 3}{-6} \cdot \left(1 - \sqrt{3}\right)^n 

17. Find an explicit formula for the sequence of exercise 39 in Section 5.6.

The recurrence relation found in exercise 39 is:

 c_k = c_{k - 1} + c_{k - 2} 

 t^k = t^{k - 1} + t^{k - 2} 

 t^2 = t^{2 - 1} + t^{2 - 2} 

 t^2 = t^1 + t^0 

 t^2 = t + 1 

 t^2 - t - 1 = 0 

 t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} 

 t = \frac{1 \pm \sqrt{5}}{2} 

 c_n = Cr^n + Ds^n 

 c_n = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n 

 c_1 = 1 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^1 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^1 

 c_2 = 2 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^2 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^2 

Let \phi = \dfrac{1 + \sqrt{5}}{2} and \psi = \dfrac{1 - \sqrt{5}}{2}. Then:

 c_1 = 1 = C \cdot \phi + D \cdot \psi 

 c_2 = 2 = C \cdot \phi^2 + D \cdot \psi^2 

 C \cdot \phi + D \cdot \psi = 1 

 C\phi = 1 - D\psi 

 C = \frac{1 - D\psi}{\phi} 

 C \cdot \phi^2 + D \cdot \psi^2 = 2 

 \frac{1 - D\psi}{\phi} \cdot \phi^2 + D \cdot \psi^2 = 2 

 (1 - D\psi) \cdot \phi + D \cdot \psi^2 = 2 

 \phi - D\psi\phi + D\psi^2 = 2 

 -D\psi\phi + D\psi^2 = 2 - \phi 

 D(-\psi\phi + \psi^2) = 2 - \phi 

 D = \frac{2 - \phi}{-\psi\phi + \psi^2} 

 D = \frac{1}{\psi}\frac{2 - \phi}{-\phi + \psi} 

Now back-substitution yields:

 D = \frac{1}{\dfrac{1 - \sqrt{5}}{2}} \cdot \frac{2 - \dfrac{1 + \sqrt{5}}{2}}{-\dfrac{1 + \sqrt{5}}{2} + \left(\dfrac{1 - \sqrt{5}}{2}\right)} 

 D = \frac{2}{1 - \sqrt{5}} \cdot \frac{\dfrac{4 - (1 + \sqrt{5})}{2}}{\dfrac{-(1 + \sqrt{5}) + (1 - \sqrt{5})}{2}} 

 D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-1 - \sqrt{5} + 1 - \sqrt{5}} 

 D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-2\sqrt{5}} 

 D = \frac{2(3 - \sqrt{5})}{(1 - \sqrt{5})(-2\sqrt{5})} 

 D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (-2\sqrt{5})(-\sqrt{5})} 

 D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (2 \cdot 5)} 

 D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + 10} 

 D = \frac{3 - \sqrt{5}}{-\sqrt{5} + 10} 

 D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} 

 D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} \cdot \frac{10 + \sqrt{5}}{10 + \sqrt{5}} 

 D = \frac{(3 - \sqrt{5})(10 + \sqrt{5})}{(10 - \sqrt{5})(10 + \sqrt{5})} 

 D = \frac{(3)(10) + (-\sqrt{5})(10) + (3)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})}{(10)(10) + (-\sqrt{5})(10) + (10)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})} 

 D = \frac{30 - 10\sqrt{5} + 3\sqrt{5} - 5}{100 - 10\sqrt{5} + 10\sqrt{5} - 5} 

 D = \frac{25 - 7\sqrt{5}}{95} 

 C = \frac{1 - \left(\dfrac{25 - 7\sqrt{5}}{95}\right)\left(\dfrac{1 - \sqrt{5}}{2}\right)}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{1 - \left(\dfrac{(25 - 7\sqrt{5})(1 - \sqrt{5})}{190}\right)}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{190 - (25 - 7\sqrt{5})(1 - \sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{190 - ((25)(1) + (-7\sqrt{5})(1) + (25)(-\sqrt{5}) + (-7\sqrt{5})(-\sqrt{5}))}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{190 - (25 - 7\sqrt{5} - 25\sqrt{5} + 35)}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{190 - (60 - 32\sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{190 - 60 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{\dfrac{130 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} 

 C = \frac{130 + 32\sqrt{5}}{190} \cdot \frac{2}{1 + \sqrt{5}} 

 C = \frac{(130 + 32\sqrt{5})(2)}{(190)(1 + \sqrt{5})} 

 C = \frac{(130 + 32\sqrt{5})(2)}{2(95)(1 + \sqrt{5})} 

 C = \frac{130 + 32\sqrt{5}}{95(1 + \sqrt{5})} 

 C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} 

 C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} 

 C = \frac{(130 + 32\sqrt{5})(1 - \sqrt{5})}{(95 + 95\sqrt{5})(1 - \sqrt{5})} 

 C = \frac{(130)(1) + (32\sqrt{5})(1) + (130)(-\sqrt{5}) + (32\sqrt{5})(-\sqrt{5})}{(95)(1) + (95\sqrt{5})(1) + (95)(-\sqrt{5}) + (95\sqrt{5})(-\sqrt{5})} 

 C = \frac{130 + 32\sqrt{5} - 130\sqrt{5} - 160}{95 + 95\sqrt{5} - 95\sqrt{5} - 475} 

 C = \frac{98\sqrt{5} - 30}{-380} 

 C = \frac{30 - 98\sqrt{5}}{380} 

 C = \frac{15 - 49\sqrt{5}}{190} 

 c_n = \frac{15 - 49\sqrt{5}}{190} \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + \frac{25 - 7\sqrt{5}}{95} \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n 

18. Suppose that the sequences s_0, s_1, s_2, \dots and t_0, t_1, t_2, \dots both satisfy the same second-order linear homogeneous recurrence relation with constant coefficients:

 s_k = 5s_{k - 1} - 4s_{k - 2} \quad \text{ for each integer } k \geq 2 

 t_k = 5t_{k - 1} - 4t_{k - 2} \quad \text{ for each integer } k \geq 2 

Show that the sequence 2s_0 + 3t_0, 2s_1 + 3t_1, 2s_2 + 3t_2, \dots also satisfies the same relation. In other words, show that

 2s_k + 3t_k = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) 

for each integer k \geq 2. Do not use Lemma 5.8.2.

Proof:

 2s_k + 3t_k 

By substitution of the given recurrence relations:

 2s_k + 3t_k = 2(5s_{k - 1} - 4s_{k - 2}) + 3(5t_{k - 1} - 4t_{k - 2}) 

 = 10s_{k - 1} - 8s_{k - 2} + 15t_{k - 1} - 12t_{k - 2} 

 = 10s_{k - 1} + 15t_{k - 1} - 8s_{k - 2} - 12t_{k - 2} 

 = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) 

This is what was to be shown.

Q.E.D.

19. Show that if r, s, a_0, and a_1 are numbers with r \neq s, then there exist unique numbers C and D so that

 C + D = a_0 

 Cr + Ds = a_1 

Proof:

Suppose r, s, a_0, and a_1 are numbers with r \neq s.

Consider the system of equations:

 C + D = a_0 

 Cr + Ds = a_1 

By solving for D and substituting:

 D = a_0 - C 

 Cr + (a_0 - C)s = a_1 

 Cr + a_0s - Cs = a_1 

 Cr - Cs = a_1 - a_0s 

 C(r - s) = a_1 - a_0s 

Since r \neq s, both sides can be divided by (r - s):

 C = \frac{a_1 - a_0s}{r - s} 

Then:

 D = a_0 - \frac{a_1 - a_0s}{r - s} 

 D = \frac{a_0(r - s) - (a_1 - a_0s)}{r - s} 

 D = \frac{a_0r - a_0s - a_1 + a_0s}{r - s} 

 D = \frac{a_0r - a_1}{r - s} 

Since r \neq s, division by r - s is valid, yielding a unique value for C, which in turn yields a unique value for D.

Q.E.D.

20. Show that if r is a nonzero real number, k and m are distinct integers, and a_k and a_m are any real numbers, then there exist unique real numbers C and D so that

 Cr^k + kDr^k = a_k 

 Cr^m + mDr^m = a_m 

Proof:

Suppose r is a nonzero real number, k and m are distinct integers, and a_k and a_m are any real numbers.

Consider the system of equations:

 Cr^k + kDr^k = a_k 

 Cr^m + mDr^m = a_m 

By solving for D and substituting:

 kDr^k = a_k - Cr^k 

Since r is a nonzero real number, we can divide by r^k:

 D = \frac{a_k - Cr^k}{kr^k} 

Then:

 Cr^m + m \cdot \frac{a_k - Cr^k}{kr^k} \cdot r^m = a_m 

 Cr^m + \frac{mr^m(a_k - Cr^k)}{kr^k} = a_m 

 \frac{Cr^m(kr^k) + mr^m(a_k - Cr^k)}{kr^k} = a_m 

 \frac{Ckr^{k + m} + mr^ma_k - mr^mCr^k}{kr^k} = a_m 

 \frac{Ckr^{k + m} - Cmr^mr^k + mr^ma_k}{kr^k} = a_m 

 \frac{Ckr^{k + m} - Cmr^{k + m} + mr^ma_k}{kr^k} = a_m 

 \frac{Cr^{k + m}(k - m) + mr^ma_k}{kr^k} = a_m 

 Cr^{k + m}(k - m) + mr^ma_k = a_m(kr^k) 

 Cr^{k + m}(k - m) = a_m(kr^k) - mr^ma_k 

 C = \frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)} 

Now we solve for D:

 D = \frac{a_k - \left(\frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)}\right)r^k}{kr^k} 

 D = \frac{a_k - \left(\frac{(a_m(kr^k) - mr^ma_k)r^k}{r^{k + m}(k - m)}\right)}{kr^k} 

 D = \frac{a_k - \left(\frac{a_m(kr^k)r^k - mr^ma_kr^k}{r^{k + m}(k - m)}\right)}{kr^k} 

 D = \frac{a_k - \left(\frac{a_mkr^{2k} - mr^{k + m}a_k}{r^{k + m}(k - m)}\right)}{kr^k} 

 D = \frac{\frac{a_k(r^{k + m}(k - m)) - (a_mkr^{2k} - mr^{k + m}a_k)}{r^{k + m}(k - m)}}{kr^k} 

 D = \frac{r^k(a_kr^k(k - m) - a_mkr^k + mr^ma_k)}{r^{k + m}(k - m)kr^k} 

 D = \frac{a_kr^k(k - m) - a_mkr^k + mr^ma_k}{r^{k + m}(k - m)k} 

 D = \frac{r^k(a_k(k - m) - a_mk) + mr^ma_k}{r^{k + m}(k - m)k} 

 D = \frac{r^k(a_k(k - m) - a_mk)}{r^{k + m}(k - m)k} + \frac{mr^ma_k}{r^{k + m}(k - m)k} 

 D = \frac{r^k(a_k(k - m) - a_mk)}{r^k(r^m(k - m)k)} + \frac{mr^ma_k}{r^m(r^k(k - m)k)} 

 D = \frac{a_k(k - m) - a_mk}{r^m(k - m)k} + \frac{ma_k}{r^k(k - m)k} 

 D = \frac{(a_k(k - m) - a_mk)(r^k) + (ma_k)(r^m)}{r^{m + k}(k - m)k} 

 D = \frac{a_k(k - m)r^k - a_mk(r^k) + ma_kr^m}{r^{m + k}(k - m)k} 

 D = \frac{a_k[(k - m)r^k + mr^m] - a_mk(r^k)}{r^{m + k}(k - m)k} 

Since r \neq 0 and k and m are distinct integers, division by r^{m + k}(k - m) is valid, yielding a unique value for C, which in turn yields a unique value for D.

Q.E.D.

21. Prove Theorem 5.8.5 for the case where the values of C and D are determined by a_0 and a_1.

Theorem 5.8.5 Single-Root Theorem

Suppose a sequence a_0, a_1, a_2, \dots satisfies a recurrence relation

 a_k = Aa_{k - 1} + Ba_{k - 2} 

for some real numbers A and B with B \neq 0 and for every integer k \geq 2. If the characteristic equation t^2 - At - B = 0 has a single (real) root r, then a_0, a_1, a_2, \dots is given by the explicit formula

 a_n = Cr^n + Dnr^n 

where C and D are the real numbers whose values are determined by the values of a_0 and any other known value of the sequence.

Proof (by strong mathematical induction):

Suppose a sequence a_0, a_1, a_2, \dots satisfies a recurrence relation

 a_k = Aa_{k - 1} + Ba_{k - 2} 

for some real numbers A and B with B \neq 0 and for every integer k \geq 2. Assume the characteristic equation t^2 - At - B = 0 has a single (real) root r.

It follows then that the sequence a_0, a_1, a_2, \dots is given by the explicit formula

 a_n = Cr^n + Dnr^n 

where C and D are the real numbers whose values are determined by the values of a_0 and any other known value of the sequence.

Let P(n) be the recurrence relation:

 a_n = Cr^n + Dnr^n 

Basis Step:

Prove P(0) and P(1), that is:

 a_0 = Cr^0 + D(0)r^0 = C 

 a_1 = Cr^1 + D(1)r^1 = Cr + Dr 

Therefore P(0) and P(1) are true by the assumption that C and D are determined by the values of a_0 and a_1.

Inductive Step:

Let k be any integer where k \geq 2. Let i be some integer such that 0 \leq i \leq k.

Suppose P(i), that is:

 a_i = Cr^i + Dir^i 

Prove P(k + 1), that is:

 a_{k + 1} = Cr^{k + 1} + D(k + 1)r^{k + 1} 

By the definition of a recurrence relation:

 a_{k + 1} = Aa_k + Ba_{k - 1} 

By the inductive hypothesis:

 = A(Cr^k + Dkr^k) + B(Cr^{k - 1} + D(k - 1)r^{k - 1}) 

By algebra:

 = ACr^k + ADkr^k + BCr^{k - 1} + BD(k - 1)r^{k - 1} 

 = ACr^k + BCr^{k - 1} + ADkr^k + BD(k - 1)r^{k - 1} 

 = C(Ar^k + Br^{k - 1}) + D(Akr^k + B(k - 1)r^{k - 1}) 

By Lemma 5.8.4, we know that because r satisfies the characteristic equation, it follows that 1, r, r^2, \dots, r^n, \dots and 0, r, 2r^2, \dots, nr^n, \dots also satisfy the same recurrence relation as a_k.

Therefore:

 = Cr^{k + 1} + D(k + 1)r^{k + 1} 

This is what was to be shown.

Q.E.D.

Exercises 22 and 23 are intended for students who are familiar with complex numbers.

22. Find an explicit formula for a sequence a_0, a_1, a_2, \dots that satisfies

 a_k = 2a_{k - 1} - 2a_{k - 1} \quad \text{ for every integer } k \geq 2 

with initial conditions a_0 = 1 and a_1 = 2.

Omitted.

23. Find an explicit formula for a sequence b_0, b_1, b_2, \dots that satisfies

 b_k = 2b_{k - 1} - 5b_{k - 2} \quad \text{ for each integer } k \geq 2 

with initial conditions b_0 = 1 and b_1 = 1.

Omitted.

24. The numbers \dfrac{1 + \sqrt{5}}{2} and \dfrac{1 - \sqrt{5}}{2} that appear in the explicit formula for the Fibonacci sequence are related to a quantity called the golden ratio in Greek mathematics. Consider a rectangle of length \phi units and height 1, where \phi > 1.

See page 387 for picture.

Divide the rectangle into a rectangle and a square as shown in the preceding diagram. The square is 1 unit on each side, and the rectangle has sides of length 1 and \phi - 1. The ancient Greeks considered the outer rectangle to be perfectly proportioned (saying that the lengths of its sides are in a golden ratio to each other) if the ratio of the length to the width of the outer rectangle equals the ratio of the length to the width of the inner rectangle. That is, if the number \phi satisfies the equation

 \frac{\phi}{1} = \frac{1}{\phi - 1} 

a. Show that if \phi satisfies the equation above, then it also satisfies the quadratic equation: t^2 - t - 1 = 0.

Omitted.

b. Find the two solutions of t^2 - t - 1 = 0 and call them \phi_1 and \phi_2.

Omitted.

c. Express the explicit formula for the Fibonacci sequence in terms of \phi_1 and \phi_2.

Omitted.