Chapter 12.5, Problem 9E

To determine

To Prove: $1^3+2^3+3^3+\mathrm{.......}+n^3=\frac{n^2{\left(n+1\right)}^2}{4}$ is True for all natural numbers $n$ .

Explanation of Solution

Given information:

Use Mathematical induction to prove the above mentioned formula.

Concept used:

1.By using Mathematical induction technique we need to prove that this formula holds True for all natural numbers. The first step is to prove True for $n=1$ .

2.Second step by using induction hypothesis of mathematical induction we assume for $n=k$ given formula is True.

3.Third step is we need to prove that above formula is True for $n=k+1$ by using formula is True for $n=k$ .

Proof:

First we need to prove the above formula is True for $n=1$ .

Given formula is

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

Put $n=1$

We get

Left hand side = $1^3=1$

Right hand side = $\frac{1^2{\left(1+1\right)}^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1$

So Left hand side = Right hand side

It True for $n=1$ .

Let us assume it is True for $n=k$

So

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

We need to prove it is True for $n=k+1$

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

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

Taking ${\left(k+1\right)}^2$ common from both terms

We get

  $\Rightarrow$ Left hand side = ${\left(k+1\right)}^2\left[\frac{k^2}{4}+\left(k+1\right)\right]$

  $\Rightarrow$ Left hand side = ${\left(k+1\right)}^2\left[\frac{k^2+4\left(k+1\right)}{4}\right]$

  $\Rightarrow$ Left hand side = ${\left(k+1\right)}^2\left[\frac{k^2+4k+4}{4}\right]$

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

  $\Rightarrow$ Right hand side = $\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2}{4}$

  $\Rightarrow$ Right hand side = ${\left(k+1\right)}^2\left[\frac{k^2+4k+4}{4}\right]$

Left hand side = Right hand side

And hence the proof by Mathematical induction.