Question

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

Answer 1

$f\left(1\right)={\left(\frac{1\times 2}{2}\right)}^2=1=1\left(true\right)$

$f\left(2\right)={\left(\frac{2\times 3}{2}\right)}^2=9=1+8\left(true\right)$

Let f(n) is true

We now, show for n=n+1

$f(n+1)=\frac{n^2{(n+1)}^2}{2}+(n+1{)}^3=f(n+1)$

$=(n+1{)}^2\left[\frac{n^2}{4}+(n+1)\right]$

$=(n+1{)}^2\left[\frac{n^2+4n+4}{4}\right]$

$=(n+1{)}^2\left[\frac{{(n+2)}^2}{4}\right]$

$=\frac{1}{4}(n+1{)}^2(n+2{)}^2$

$=\frac{1}{4}\left[\right(n+1\left)\right(n+2){]}^2$

Hence Proved