Question

Solve
$1^2+(1^2+2^2)+(1^2+2^2+3^2)$

Answer 1

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

For general term,

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

$={\sum }_{i=1}^n{\sum }_{0=1}^i{j}^2$

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

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

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

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

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

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

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

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

So, general term$=\frac{n(n+1{)}^2(n+2)}{12}$

Now, for $n=3$,

$\frac{3\left(4{)}^2\right(5)}{12}$

$=\frac{3\times 16\times 5}{12}=20$