Chapter 11.4, Problem 8E

To determine

To prove:

Given statement is true using principle of mathematical induction.

Explanation of Solution

Give information:

  $1^3+2^3+3^3+\mathrm{...}+n^3={\left(1+2+3+\mathrm{...}+n\right)}^2$

Formula Used:

Principle of mathematical Induction.

Proof:

It is given that,

  $1^3+2^3+3^3+\mathrm{...}+n^3={\left(1+2+3+\mathrm{...}+n\right)}^2$

For $n=1$ the equation becomes,

  $\begin{array}{l}{n}^3=n^2\\ 1^3=1^{\begin{array}{l}2\\ 1\end{array}}\\ 1=1\end{array}$

Thus, statement is true for $n=1$

Assume that the statement is true for $n=k$

  $1^3+2^3+3^3+\mathrm{...}+k^3={\left(1+2+3+\mathrm{...}+k\right)}^2$

Is true for all values of $k$ ,

Prove that statement is true for $n=k+1$ ,

So,

  $\begin{array}{l}{1}^3+2^3+3^3+\mathrm{...}+n^3={\left(1+2+3+\mathrm{...}+n\right)}^2\\ 1^3+2^3+3^3+\mathrm{...}+{\left(k+1\right)}^3={\left(1+2+3+\mathrm{...}+\left(k+1\right)\right)}^2\end{array}$

Since, $1^3+2^3+3^3+\mathrm{...}+k^3={\left(1+2+3+\mathrm{...}+k\right)}^2$

The above statement becomes,

  $\begin{array}{l}\left(\frac{k{\left(k+1\right)}^2}{2}\right)+{\left(k+1\right)}^3={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\\ \frac{k^2{\left(k+1\right)}^2}{4}+{\left(k+1\right)}^3={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\end{array}$

Take least common multiple to the left hand side,

  $\left(\frac{k^2{\left(k+1\right)}^2}{4}+{\left(k+1\right)}^3\right)={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2$

Take common term outside the other terms,

  $\begin{array}{l}\left(\frac{{\left(k+1\right)}^2\left(k^2+4\left(k+1\right)\right)}{4}\right)={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\\ \left(\frac{{\left(k+1\right)}^2\left(k^2+4k+4\right)}{4}\right)={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\end{array}$

Factorize the second term in the numerator of the left hand side,

  $\begin{array}{l}\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2}{4}={\left(\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2}{2}\right)}^2\\ \left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\end{array}$

Thus left hand side is equal to the right hand side.

Therefore, the statement is true for all positive integers $n$ .

Therefore, by principle of mathematical induction $P\left(n\right)$ is true for all positive integers $n$