Chapter 12.5, Problem 11E

To determine

To Prove: $2^3+4^3+6^3+\mathrm{......}+{\left(2n\right)}^3=2n^2{\left(n+1\right)}^2$ 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

  $2^3+4^3+6^3+\mathrm{......}+{\left(2n\right)}^3=2n^2{\left(n+1\right)}^2$

Put $n=1$

We get

Left hand side = $2^3=8$

Right hand side = $2{\left(1\right)}^2{\left(1+1\right)}^2=2{\left(2\right)}^2=8$

So Left hand side = Right hand side

It True for $n=1$ .

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

So

  $2^3+4^3+6^3+\mathrm{......}+{\left(2k\right)}^3=2k^2{\left(k+1\right)}^2$ is True

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

Left hand side = $2^3+4^3+6^3+\mathrm{......}+{\left(2k\right)}^3+{\left(2\left(k+1\right)\right)}^3$

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

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

We get

  $\Rightarrow$ Left hand side = ${\left(k+1\right)}^2\left[2k^2+8\left(k+1\right)\right]$

  $\Rightarrow$ Left hand side = ${\left(k+1\right)}^2\left[2k^2+8k+8\right]$

Taking 2 common from terms of second term

We get

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

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

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

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

Left hand side = Right hand side

And hence the proof by Mathematical induction.