Question

Find the sum of n terms of the series whose nth term is $n^2(n^2-1)$.

Answer 1

$nth$ term is $n^4-n^2$

Sum of $n$ terms is $\Sigma (n^4-n^2)=\Sigma n^4-\Sigma n^2$

now we know $\Sigma n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

and $\Sigma n^2=\frac{n(n+1)(2n+1)}{6}$

So, $\Sigma (n^4-n^2)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}-\frac{n(n+1)(2n+1)}{6}$

$\Rightarrow \Sigma (n^4-n^2)=\frac{n(n+1)(2n+1)(n^2+n-2)}{10}$

$\Rightarrow \Sigma (n^4-n^2)=\frac{n(n+1)(2n+1)(n+2)(n-1)}{10}$