Question

Show that $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$ upto $n$ terms $=\frac{n(n+1{)}^2(n+2)}{12},\forall n\in N$

Answer 1

Formuale being used:

${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$

${\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$

${\sum }_{i=1}^n{i}^3=\frac{n^2(n+1{)}^2}{4}$

Given,

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+..............+(1^2+2^2+3^2+........+n^2)$

$={\sum }_{i=1}^n(1^2+2^2+3^2+........+i^2)$

$={\sum }_{i=1}^n{\sum }_{j=1}^i{j}^2$

$={\sum }_{i=1}^n\frac{i(i+1)(2i+1)}{6}$

$={\sum }_{i=1}^n\frac{2i^3+3i^2+i}{6}$

$={\sum }_{i=1}^n\frac{1}{3}{i}^3+{\sum }_{j=1}^n\frac{1}{2}{j}^2+{\sum }_{k=1}^n\frac{1}{6}k$

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

$=\frac{n^2(n+1{)}^2+n(n+1\left)\right(2n+1)+n(n+1)}{12}$

$=\frac{n(n+1)\left[n\right(n+1)+(2n+1)+1]}{12}$

$=\frac{n(n+1)(n^2+3n+2)}{12}$

$=\frac{n(n+1{)}^2(n+2)}{12}$