Question

Find the sum to n terms of the series $S_n=1^2+(1^2+2^2)+(1^2+2^2+3^2)....Also,determine\sum \frac{S_n}{(n+1{)}^2}$.

Answer 1

Let $a_n$ be the nthe term of the given series then

$a_n=1^2+2^2+3^2+...n^2=\frac{s(n+1)(2n+1)}{6}=\frac{1}{6}(2n^3+3n^2+n)$

Let $S_n$ be the sum to n terms of the given series , then

$S_n=\sum a_n=\sum \frac{1}{6}(2n^3+3n^2+n)$

$\Rightarrow S_n=\frac{2}{6}\sum n^3+\frac{3}{6}\sum n^2+\frac{1}{6}\sum n$

$\Rightarrow S_n=\frac{2}{6}{\left[\frac{n(n+1)}{2}\right]}^2+\frac{3}{6}\times \frac{n(n+1)(2n+1)}{6}+\frac{1}{6}.\frac{n(n+1)}{2}$

$\Rightarrow S_n=\frac{n(n+1)}{12}\left[n\right(n+1)+(2n+1+1\left)\right]\Rightarrow S_n=\frac{n(n+1)}{12}(n^2+3n+2)$

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

$\therefore \sum \frac{S_n}{(n+1{)}^2}=\sum \frac{n(n+1{)}^2(n+2)}{12\times (n+1{)}^2}\sum \frac{n(n+2)}{12}=\sum \frac{n^2+2n}{12}$

$=\frac{1}{12}[\sum n^2+2\sum n]$

$=\frac{1}{12}[\frac{n(n+1)(2n+1)}{6}+2\times n\frac{(n+1)}{2}]=\frac{n(n+1)}{12}=[\frac{(2n+1)}{6}+1]$

$=\frac{n(n+1)(2n+7)}{72}$