Question

Find the sum to n terms of the series whose nth term is given by

$n(n+1)(n+4)$

Answer 1

Let $a_n$ be the nth term of the given series and $S_n$ be the sum of n terms of the given series.

$\therefore a_n=n(n+1)(n+4)$ [Given]

$=n^3+5n^2+4n$

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(k^3+5k^2+4k)$

$=(1^3+{5.1}^2+4.1)+(2^3+{5.2}^2+4.2)+\dots \dots +(n^3+5n^2+4n)$

$=(1^3+2^3+\dots \dots +n^3)+5(1^2+2^2+3^2+\dots \dots +n^2)+4(1+2+3+\dots \dots +n)$

$=\frac{\left[n\right(n+1){]}^2}{2}+\frac{5n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}$

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

$=n(n+1)[\frac{n(n+1)}{4}+\frac{5(2n+1)}{6}+2]$

$=n(n+1)\left[\frac{3n^2+3n+20n+10+2n}{12}\right]$

$=\frac{n(n+1)(3n^2+23n+34)}{12}$