Chapter 5 - Section 5.1 - Mathematical Induction - Exercises - Page 329: 4

Let $ P (n)$ be the proposition that: $$ 1^{3}+2^{3}+3^{3}+\cdots+( n)^{3}=\left[\frac{n(n+1)}{2}\right]^{2} $$ (a) by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement $$ 1^{3}=\left[\frac{1(1+1)}{2}\right]^{2} $$ (b) $P(1)$ is true, because $$ 1^{3}=\left[\frac{1(1+1)}{2}\right]^{2}=1 $$ (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 $\left[\frac{n(n+1)}{2}\right]^{2}$) (c) For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer $k$. That is, we assume that: $$ 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}=\left[\frac{k(k+1)}{2}\right]^{2} $$ (d) Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{3}+2^{3}+3^{3}+\cdots+( k+1)^{3}=\left[\frac{(k+1)(k+2)}{2}\right]^{2} $$ 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 $$ 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}=\left[\frac{k(k+1)}{2}\right]^{2} $$ Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{3}+2^{3}+3^{3}+\cdots+( k+1)^{3}=\left[\frac{(k+1)(k+2)}{2}\right]^{2} $$ is also true. When we add $( k + 1)^{3}$ to both sides of the equation in $P(k)$, we obtain $$ \begin{split} 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}+ ( k + 1)^{3}&= \\ & \quad[\text { by the inductive hypothesis }] \\ &=\left[\frac{k(k+1)}{2}\right]^{2}+( k + 1)^{3}\\ &=\frac{(k+1)^{2}}{4}(k^{2}+4(k+1)) \\ &=\frac{(k+1)^{2}}{4}\left( k^{2}+4 k+4\right)\\ &=\frac{(k+1)^{2}}{4}(k+2)^{2}\\ &=\left[\frac{(k+1)(k+2)}{2}\right]^{2} \end{split} $$ (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: $$ 1^{3}+2^{3}+3^{3}+\cdots+( n)^{3}=\left[\frac{n(n+1)}{2}\right]^{2} $$ (a) by substituting $n=1 $ in the proposition $ P (n)$ , we obtain that $P (1)$ is the statement $$ 1^{3}=\left[\frac{1(1+1)}{2}\right]^{2} $$ (b) $P(1)$ is true, because $$ 1^{3}=\left[\frac{1(1+1)}{2}\right]^{2}=1 $$ (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 $\left[\frac{n(n+1)}{2}\right]^{2}$) (c) For the inductive hypothesis we assume that P(k) holds for an arbitrary positive integer $k$. That is, we assume that: $$ 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}=\left[\frac{k(k+1)}{2}\right]^{2} $$ (d) Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{3}+2^{3}+3^{3}+\cdots+( k+1)^{3}=\left[\frac{(k+1)(k+2)}{2}\right]^{2} $$ 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 $$ 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}=\left[\frac{k(k+1)}{2}\right]^{2} $$ Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that $$ 1^{3}+2^{3}+3^{3}+\cdots+( k+1)^{3}=\left[\frac{(k+1)(k+2)}{2}\right]^{2} $$ is also true. When we add $( k + 1)^{3}$ to both sides of the equation in $P(k)$, we obtain $$ \begin{split} 1^{3}+2^{3}+3^{3}+\cdots+( k)^{3}+ ( k + 1)^{3}&= \\ & \quad[\text { by the inductive hypothesis }] \\ &=\left[\frac{k(k+1)}{2}\right]^{2}+( k + 1)^{3}\\ &=\frac{(k+1)^{2}}{4}(k^{2}+4(k+1)) \\ &=\frac{(k+1)^{2}}{4}\left( k^{2}+4 k+4\right)\\ &=\frac{(k+1)^{2}}{4}(k+2)^{2}\\ &=\left[\frac{(k+1)(k+2)}{2}\right]^{2} \end{split} $$ (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 $n$.