Question

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

Answer 1

Given!

$n^{th}$term $=n^2(n^2-1)$

Sum of $n$ term is denoted by $S_n$

Then, $S_n=\sum _{k=1}^n{k}^4-k^2$

$=\sum _{k=1}^n{k}^4-\sum _{k=1}^n{n}^2$

$=\{\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}\}-\frac{n(n+1)(2n+2)}{6}$

$=\frac{n^5}{5}+\frac{n^4}{2}-\frac{2n^2}{3}-\frac{11n}{30}$