Chapter 5 - Section 5.1 - Mathematical Induction - Exercises - Page 329: 3

Let $ P (n)$ be the proposition that: $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ (a) by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement $$ 1^{2}=\frac{1(1+1)(2(1)+1)}{6} $$ (b) $P(1)$ is true, because $$ 1^{2}=\frac{1(1+1)(2(1)+1)}{6}=\frac{(2)(3)}{6}=1 $$ (The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for $n$ in $\frac{n(n+1)(2n+1)}{6}$) (c) For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that: $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ (d) Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k+1)^{2}=\frac{(k+1)(k+2)(2k+3)}{6} $$ is also true (e) For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k$. That is, we assume that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k+1)^{2}=\frac{(k+1)(k+2)(2k+3)}{6} $$ is also true. When we add $( k + 1)^{2}$ to both sides of the equation in $P(k)$, we obtain $$ \begin{split} 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}+ ( k + 1)^{2}&= \\ & \quad[\text { by the inductive hypothesis }] \\ &=\frac{k(k+1)(2k+1)}{6}+( k + 1)^{2}\\ &=\frac{k+1}{6}(k(2 k+1)+6(k+1)) \\ &=\frac{k+1}{6}\left(2 k^{2}+7 k+6\right)\\ &=\frac{k+1}{6}(k+2)(2 k+3)\\ &=\frac{(k+1)(k+2)(2 k+3)}{6} \end{split} $$ (f) We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.

Let $ P (n)$ be the proposition that: $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ (a) by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement $$ 1^{2}=\frac{1(1+1)(2(1)+1)}{6} $$ (b) $P(1)$ is true, because $$ 1^{2}=\frac{1(1+1)(2(1)+1)}{6}=\frac{(2)(3)}{6}=1 $$ (The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for $n$ in $\frac{n(n+1)(2n+1)}{6}$) (c) For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer k. That is, we assume that: $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ (d) Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k+1)^{2}=\frac{(k+1)(k+2)(2k+3)}{6} $$ is also true (e) For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k$. That is, we assume that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}=\frac{k(k+1)(2k+1)}{6} $$ Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{2}+2^{2}+3^{2}+\cdots+( k+1)^{2}=\frac{(k+1)(k+2)(2k+3)}{6} $$ is also true. When we add $( k + 1)^{2}$ to both sides of the equation in $P(k)$, we obtain $$ \begin{split} 1^{2}+2^{2}+3^{2}+\cdots+( k)^{2}+ ( k + 1)^{2}&= \\ & \quad[\text { by the inductive hypothesis }] \\ &=\frac{k(k+1)(2k+1)}{6}+( k + 1)^{2}\\ &=\frac{k+1}{6}(k(2 k+1)+6(k+1)) \\ &=\frac{k+1}{6}\left(2 k^{2}+7 k+6\right)\\ &=\frac{k+1}{6}(k+2)(2 k+3)\\ &=\frac{(k+1)(k+2)(2 k+3)}{6} \end{split} $$ (f) We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.