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@ -7152,36 +7152,144 @@ 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 $k \geq 2$ $a_1 = 1$
2. $b_k = b_{k - 1} + 3_k$, for every integer $k \geq 2$ $b_1 = 1$
$$
a_1 = 1 \\
a_2 = 2a_1 + 2 = 2(1) + 2 = 2 + 2 = 4 \\
a_3 = 2a_2 + 3 = 2(4) + 3 = 8 + 3 = 11 \\
a_4 = 2a_3 + 4 = 2(11) + 4 = 22 + 4 = 26
$$
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 $k \geq 1$.
10. let $b_0, b_1, b_2, \dots$ be defined by the formula $b_n = 4^n$, for every
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
@ -7189,27 +7297,159 @@ Find the first four terms of each of the recursively defined sequences in 1-8.
$$ s_k = \frac{-s_{k - 1}}{k} $$
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 $n \geq 1$,
$$ C_n = \frac{1}{4n + 2}\binom{2n + 2}{n + 1}$$
$$ C_n = \frac{1}{4n + 2}\binom{2n + 2}{n + 1} $$
_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
@ -7222,10 +7462,42 @@ $$ a_n = \left[\text{the minimum number of moves needed to transfer a tower of }
a. Find $a_1, a_2$, and $a_3$.
$$ 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 $$
18. _Tower of Hanoi with Adjacency Requirement:_
Suppose the same situation as in exercise 17. Let
@ -7234,15 +7506,102 @@ $$ b_n = \left[\text{the minimum number of moves needed to transfer a tower of }
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

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@ -152,16 +152,26 @@ Page 359
1. A recursive definition for a sequence consists of a _____ and _____.
recurrence relation; initial conditions
2. A recurrence relation is an equation that defines each later term of a
sequence by reference to _____ in the sequence.
earlier terms
3. Initial conditions for a recursive definition of a sequence consist of one or
more of the _____ of the sequence.
values of the first few terms
4. To solve a problem recursively means to divide the problem into smaller
subproblems of the same type as the initial problem, to suppose _____, and to
figure out how to use the supposition to _____.
that the smaller subproblems have already been solved; solve the initial problem
5. A crucial step for solving a problem recursively is to define a _____ in
terms of which the recurrence relation and initial conditions can be
specified.
sequence

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