Question

Find the sum of $n$ terms of $1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+...$ from that find the sum of the first $10$ terms

Answer 1

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

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

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

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

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

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

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

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

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

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

$\therefore S_{10}=\frac{10(10+1{)}^2(10+2)}{12}$

$=\frac{10\left(11{)}^2\right(12)}{12}$

$=1210$