✏️ Finished 5.7
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@ -10457,27 +10457,385 @@ verify that the formula of part (a) is correct.
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43. $a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1}$, for each integer $k \geq 1$
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$a_0 = 2$.
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a.
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$$ a_0 = 2 $$
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$$ a_1 = \frac{a_0}{2a_0 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} $$
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$$ a_2 = \frac{a_1}{2a_1 - 1} = \frac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} = \frac{2}{4 - 3} = 2 $$
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$$ a_3 = \frac{a_2}{2a_2 - 1} = \frac{2}{2(2) - 1} = \frac{2}{3} $$
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Guess:
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$$
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a_n =
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\begin{cases}
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2 & \text{if } n \text{ is even} \\
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\dfrac{2}{3} & n \text{ is odd}
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\end{cases}
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$$
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b.
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Let $a_0, a_1, a_2, \dots$ be the sequence defined recursively by $a_0 = 2$ and
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$a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1}$ for each integer $k \geq 1$.
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Let the property $P(n)$ be the equation:
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$$
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a_n =
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\begin{cases}
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2 & \text{if } n \text{ is even} \\
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\dfrac{2}{3} & \text{if } n \text{ is odd}
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\end{cases}
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$$
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**Proof by strong mathematical induction:**
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_Basis Step:_
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Prove $P(0)$ and $P(1)$, that is:
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$$
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a_0 =
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\begin{cases}
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2 & \text{if } 0 \text{ is even} \\
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\dfrac{2}{3} & \text{if } 0 \text{ is odd}
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\end{cases}
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$$
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and:
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$$
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a_1 =
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\begin{cases}
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2 & \text{if } 1 \text{ is even} \\
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\dfrac{2}{3} & \text{if } 1 \text{ is odd}
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\end{cases}
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$$
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$P(0)$ is true since the piecewise function tells us that $a_0 = 2$ since $0$ is
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even and this matches the given value of $a_0 = 2$.
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$P(1)$ is true given the evaluation of $a_1$ in part (a).
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Therefore both $P(0)$ and $P(1)$ are true.
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_Inductive Step:_
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Let $k$ be any integer such that $k \geq 0$. Let $i$ be some integer such that
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$0 \leq i \leq k$.
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Suppose $P(i)$, that is:
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$$
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a_i =
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\begin{cases}
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2 & \text{if } i \text{ is even} \\
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\dfrac{2}{3} & \text{if } i \text{ is odd}
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\end{cases}
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$$
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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$$
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a_{k + 1} =
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\begin{cases}
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2 & \text{if } k + 1 \text{ is even} \\
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\dfrac{2}{3} & \text{if } k + 1 \text{ is odd}
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\end{cases}
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$$
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By the definition of the sequence:
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$$ a_k = \dfrac{a_{k - 1}}{2a_{k - 1} - 1} $$
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It follows that:
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$$ a_{k + 1} = \dfrac{a_k}{2a_k - 1} $$
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By the inductive hypothesis:
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$$
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a_{k + 1} =
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\begin{cases}
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\dfrac{2}{2(2) - 1} & \text{if } k \text{ is even} \\
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\dfrac{\dfrac{2}{3}}{2\left(\dfrac{2}{3}\right) - 1} & \text{if } k \text{ is odd}
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\end{cases}
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$$
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$$
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= \\
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\begin{cases}
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\dfrac{2}{3} & \text{if } k \text{ is even} \\
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2 & \text{if } k \text{ is odd}
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\end{cases}
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$$
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It follows that:
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$$
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= \\
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\begin{cases}
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\dfrac{2}{3} & \text{if } k + 1 \text{ is odd} \\
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2 & \text{if } k + 1 \text{ is even}
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\end{cases}
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$$
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This is what was to be shown. Therefore $P(k + 1)$ is true.
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Q.E.D.
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44. $b_k = \dfrac{2}{b_{k - 1}}$, for each integer $k \geq 2$ $b_1 = 1$.
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a.
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$$ b_1 = 1 $$
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$$ b_2 = \frac{2}{b_1} = \frac{2}{1} = 2 $$
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$$ b_3 = \frac{2}{b_2} = \frac{2}{2} = 1 $$
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Guess:
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$$
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b_n =
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\begin{cases}
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1 & \text{if } n \text{ is odd} \\
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2 & \text{if } n \text{ is even}
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\end{cases}
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$$
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b.
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Let $b_1, b_2, b_3, \dots$ be the sequence defined recursively by $b_1 = 1$ and
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$b_k = \dfrac{2}{b_{k - 1}}$ for each integer $k \geq 2$.
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**Proof by strong mathematical induction:**
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Let the property $P(n)$ be the equation:
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$$
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b_n =
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\begin{cases}
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1 & \text{if } n \text{ is odd} \\
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2 & \text{if } n \text{ is even}
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\end{cases}
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$$
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_Basis Step:_
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Prove $P(1)$ and $P(2)$.
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Both $P(1)$ and $P(2)$ are proven true in part (a).
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_Inductive Step:_
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Let $k$ be any integer such that $k \geq 2$, and let $i$ be some integer such
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that $1 \leq i \leq k$.
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Suppose $P(i)$, that is:
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$$
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b_i =
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\begin{cases}
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1 & \text{if } i \text{ is odd} \\
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2 & \text{if } i \text{ is even}
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\end{cases}
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$$
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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$$
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b_{k + 1} =
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\begin{cases}
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1 & \text{if } k + 1 \text{ is odd} \\
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2 & \text{if } k + 1 \text{ is even}
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\end{cases}
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$$
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By the definition of the given sequence:
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$$ b_k = \dfrac{2}{b_{k - 1}} $$
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It follows that:
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$$ b_{k + 1} = \dfrac{2}{b_k} $$
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By the inductive hypothesis:
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$$
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b_{k + 1} =
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\begin{cases}
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\dfrac{2}{1} & \text{if } k \text{ is odd} \\
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\dfrac{2}{2} & \text{if } k \text{ is even}
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\end{cases}
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$$
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$$
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= \\
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\begin{cases}
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2 & \text{if } k \text{ is odd} \\
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1 & \text{if } k \text{ is even}
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\end{cases}
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$$
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$$
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= \\
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\begin{cases}
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2 & \text{if } k + 1 \text{ is even} \\
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1 & \text{if } k + 1 \text{ is odd}
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\end{cases}
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$$
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This is what was to be shown. Therefore $P(k + 1)$ is true.
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Q.E.D.
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45. $v_k = v_{\lfloor \dfrac{k}{2} \rfloor} + v_{\lfloor \dfrac{(k + 1)}{2}\rfloor} + 2$,
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for each integer $k \geq 2$ $v_1 = 1$.
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a.
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$$ v_1 = 1 $$
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$$ v_2 = v_{\lfloor \dfrac{2}{2} \rfloor} + v_{\lfloor \dfrac{(2 + 1)}{2} \rfloor} + 2 $$
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$$ = v_1 + v_1 + 2 = 1 + 1 + 2 $$
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$$ v_3 = v_{\lfloor \dfrac{3}{2} \rfloor} + v_{\lfloor \dfrac{(3 + 1)}{2} \rfloor} + 2 $$
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$$ = v_1 + v_2 + 2 = 1 + (1 + 1 + 2) + 2 $$
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$$ = 3 + 2 \cdot 2 $$
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$$ v_4 = v_{\lfloor \dfrac{4}{2} \rfloor} + v_{\lfloor \dfrac{(4 + 1)}{2} \rfloor} + 2 $$
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$$ = v_2 + v_2 + 2 = (1 + 1 + 2) + (1 + 1 + 2) + 2 $$
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$$ = 4 + 3 \cdot 2 $$
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$$ v_5 = v_{\lfloor \dfrac{5}{2} \rfloor} + v_{\lfloor \dfrac{(5 + 1)}{2} \rfloor} + 2 $$
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$$ = v_2 + v_3 + 2 = (1 + 1 + 2) + (3 + 2 \cdot 2) + 2 $$
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$$ = 5 + 4 \cdot 2 $$
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Guess:
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$$ v_n = n + 2(n - 1) = n + 2n - 2 $$
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$$ = 3n - 2 $$
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b.
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Let $v_1, v_2, v_3, \dots$ be the sequence defined recursively by $v_1 = 1$ and
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$v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2$
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for each integer $k \geq 2$.
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**Proof by strong mathematical induction:**
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Let the property $P(n)$ be the equation $v_n = 3n - 2$.
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_Basis Step:_
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Prove $P(1)$, that is:
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$$ v_1 = 3(1) - 2 $$
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$$ = 3 - 2 $$
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$$ = 1 $$
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This equality matches the given value of $v_1 = 1$. Therefore $P(1)$ is true.
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_Inductive Step:_
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Let $k$ be any integer such that $k \geq 2$, and let $i$ be some integer such
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that $1 \leq i \leq k$.
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Suppose $P(i)$, that is:
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$$ v_i = 3i - 2 $$
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This is the inductive hypothesis.
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Prove $P(k + 1)$, that is:
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$$ v_{k + 1} = 3(k + 1) - 2 $$
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Alternatively:
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$$ v_{k + 1} = 3k + 1 $$
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By the definition of the given sequence:
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$$ v_k = v_{\lfloor \frac{k}{2} \rfloor} + v_{\lfloor \frac{(k + 1)}{2} \rfloor} + 2 $$
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It follows that:
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$$ v_{k + 1} = v_{\lfloor \frac{(k + 1)}{2} \rfloor} + v_{\lfloor \frac{(k + 2)}{2} \rfloor} + 2 $$
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By the inductive hypothesis:
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$$ = \left(3\lfloor \frac{k + 1}{2} \rfloor - 2\right) + \left(3\lfloor \frac{k + 2}{2}\rfloor - 2\right) + 2 $$
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$$ = 3\left(\lfloor \frac{k + 1}{2} \rfloor + \lfloor \frac{k + 2}{2} \rfloor \right) - 2 $$
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$$
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= \\
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\begin{cases}
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3\left(\frac{k}{2} + \frac{k + 2}{2}\right) - 2 & \text{if } k \text{ is even} \\
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3\left(\frac{k + 1}{2} + \frac{k + 1}{2}\right) - 2 & \text{if } k \text{ is odd}
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\end{cases}
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$$
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$$ = 3\left(\frac{2k + 2}{2}\right) - 2 $$
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$$ = 3(k + 1) - 2 $$
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$$ = 3k + 3 - 2 $$
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$$ = 3k + 1 $$
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This is what was to be shown. Therefore $P(k + 1)$ is true.
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Q.E.D.
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46. $s_k = 2s_{k - 2}$, for each integer $k \geq 2$ $s_0 = 1$, $s_1 = 2$.
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Omitted.
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47. $t_k = k - t_{k - 1}$, for each integer $k \geq 1$ $t_0 = 0$.
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Omitted.
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48. $w_k = w_{k - 2} + k$, for each integer $k \geq 3$ $w_1 = 1$, $w_2 = 2$.
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Omitted.
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49. $u_k = u_{k - 2} \cdot u_{k - 1}$, for each integer $k \geq 2$
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$u_0 = u_1 = 2$
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Omitted.
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In 50 and 51 determine whether the given recursively defined sequence satisfies
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the explicit formula $a_n = (n - 1)^2$, for every integer $n \geq 1$.
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50. $a_k = 2a_{k - 1} + k - 1$, for each integer $k \geq 2$ $a_1 = 0$.
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Omitted.
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51. $a_k = 4a_{k - 1} - k + 3$, for each integer $k \geq 2$ $a_1 = 0$.
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Omitted.
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52. A single line divides a plane into two regions. Two lines (by crossing) can
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divide a plane into four regions; three lines can divide it into seven
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regions (see the figure). Let $P_n$ be the maximum number of regions into
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@ -10488,13 +10846,19 @@ the explicit formula $a_n = (n - 1)^2$, for every integer $n \geq 1$.
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a. Derive a recurrence relation for $P_k$ in terms of $P_{k - 1}$, for each
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integer $k \geq 2$.
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Omitted.
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b. Use iteration to guess an explicit formula for $P_n$.
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Omitted.
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53. Compute $\left[\begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array}\right]^n$ for
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small values of $n$ (up to about 5 or 6). Conjecture explicit formulas for
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the entries in this matrix, and prove your conjecture using mathematical
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induction.
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Omitted.
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54. In economics the behavior of an economy from one period to another is often
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modeled by recurrence relations. Let $Y_k$ be the income in period $k$ and
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$C_k$ be the consumption in period $k$. In one economic model, income in any
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@ -10516,9 +10880,13 @@ $Y_k = E + c + mY_{k - 1}$.
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a. Use iteration on the above recurrence relation to obtain
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Omitted.
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$$ Y_n = (E + c)\left(\frac{m^n - 1}{m - 1}\right) + m^nY_0 $$
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for every integer $n \geq 1$.
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b. (For students who have studied calculus) Show that if $0 < m < 1$, then
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$\lim\limits_{m \to \infty}Y_n = \dfrac{E + c}{1 - m}$.
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Omitted.
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