Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}\text{ Let }S& =\sum _{k=1}^{n}k(k+1)(k-1)\\ & =\sum _{k=1}^n{k}^3-k\end{array}$ We know that ${\sum }_{r=1}^{P}r=\frac{P(P+1)}{2}$ and ${\sum }_{r=1}^P{r}^3={\left[\frac{P(P+1)}{2}\right]}^2$ $\begin{array}{rl}S& ={\left[\frac{n(n+1)}{2}\right]}^2-\left[\frac{n(n+1)}{2}\right]\\ & =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}-1\right]\\ & =\left(\frac{n^2+n}{2}\right)\left[\frac{n^2+n-2}{2}\right]\\ & =\frac{n^4}{4}+\frac{n^3}{2}-\frac{n^2}{4}-\frac{n}{2}...(i)\end{array}$ Now, by comparing $p{n}^4+q{n}^3+t{n}^2+sn$ with Eq. (i), $S=-\frac{1}{2}$