diff --git a/appendix_b.txt b/appendix_b.txt index 2abc897..a5356d9 100644 --- a/appendix_b.txt +++ b/appendix_b.txt @@ -1 +1 @@ -933 +935 diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 35b7291..a9b29c2 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -852,49 +852,207 @@ $\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4 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 $P(1)$. Is $P(1)$ true? +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 -$n \geq 1$, what must be shown in the inductive step? +$n \geq 2$, 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 @@ -955,6 +1113,18 @@ 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. @@ -962,48 +1132,1005 @@ from Theorem 5.2.1 or Theorem 5.2.2. $$ 2 + 4 + 6 + \dots + 2n = n^2 + n $$ +**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 $n \geq 0$, -$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} + 1 $$ +$$ 1 + 2 + 2^2 + \dots + 2^n = 2^{n + 1} - 1 $$ + +**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 $n \geq 1$, diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index 490a7c7..8914245 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -40,12 +40,18 @@ Page 309 1. Mathematical induction is a method for proving that a property defined for integers $n$ is true for all values of $n$ that are _____. +greater than or equal to some initial value. + 2. Let $P(n)$ be a property defined for integers $n$ and consider constructing a proof by mathematical induction for the statement "P(n) is true for all $n \geq a$." a. In the basis step one must show _____. +that $P(a)$ is true. + b. In the inductive step one supposes that _____ for a particular but arbitrarily chosen value of an integer $k \geq a$. This supposition is called the _____. One then has to show that _____. + +$P(k)$ is true; inductive hypothesis; $P(k + 1)$ is true. diff --git a/leftoff.txt b/leftoff.txt index a1f7f63..54ea97e 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -298 +310