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@ -4843,6 +4843,47 @@ 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:
$$ b_1 = 4, b_2 = 12, b_k = b_{k - 2} + b_{k - 1} $$
@ -4851,6 +4892,54 @@ 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:
$$ c_0 = 2, c_1 = 2, c_2 = 6, c_k = 3c_{k - 3} $$
@ -4859,6 +4948,62 @@ 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:
$$ d_1 = \frac{9}{10}, d_2 = \frac{10}{11}, d_k = d_{k - 1} \cdot d_{k - 2} $$
@ -4867,6 +5012,46 @@ 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:
$$ e_0 = 12, e_1 = 29, e_k, = 5e_{k - 1} - 6e_{k - 2} $$
@ -4875,6 +5060,79 @@ 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:
$$ f_0 = 5, f_1 = 16, f_k = 7f_{k - 1} - 10f_{k - 2} $$
@ -4883,6 +5141,77 @@ 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:
$$ g_1 = 3, g_2 = 5, g_k = 3g_{k - 1} - 2g_{k - 2} $$
@ -4891,6 +5220,77 @@ 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:
$$ h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k - 1} + h_{k - 2} + h_{k - 3} $$
@ -4899,21 +5299,200 @@ 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