Chapter F, Problem 38E

To determine

To prove: The formula ${\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2$ by mathematical induction.

Explanation of Solution

Let $S_n$ be the formula ${\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2$.

Consider the formula ${\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2$ for $n=1$.

$\begin{array}{c}{\displaystyle \sum _{i=1}^1{i}^3}=1^3\\ ={\left[\frac{1\left(1+1\right)}{2}\right]}^2\\ =1\end{array}$

That is, $S_1$ is true.

Assume that that the formula $S_n$ is true for $n=k$.

That is, ${\displaystyle \sum _{i=1}^k{i}^3}={\left[\frac{k\left(k+1\right)}{2}\right]}^2$.

To show the formula is true for $n=k+1$.

$\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^3}={\displaystyle \sum _{i=1}^k{i}^3}+{\left(k+1\right)}^3\\ ={\left[\frac{k\left(k+1\right)}{2}\right]}^2+{\left(k+1\right)}^3\\ =\frac{{\left(k+1\right)}^2}{4}\left[k^2+4\left(k+1\right)\right]\\ =\frac{{\left(k+1\right)}^2}{4}{\left(k+2\right)}^2\end{array}$

Further simplified as,

$\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^3}={\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]}^2\\ ={\left(\frac{\left(k+1\right)\left[\left(k+1\right)+1\right]}{2}\right)}^2\end{array}$

Thus, the formula is true for $n=k+1$.

Therefore, the formula $S_n$ is true for all values of n by mathematical induction.

Hence the formula is proved.