Question

Find the sum to $n$ terms of the series $(n^2-1^2)+2(n^2-2^2)+3(n^2-3^2)+...$

Answer 1

$1^2(n^2-1^2)+2(n^2-2^2)+3(n^2-3^2)$….

$n^2(1+2+\mathrm{3....}n)-(1+2^3+3^3+4^3.....n^3)$

$\therefore \frac{n^2\left(n\right)(n+1)}{2}-\frac{\left(n{)}^2\right(n+1{)}^2}{4}$

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

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

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

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

$=\frac{n^2}{4}(n^2-1)$

Answer: $\left(\frac{n^4-n^2}{4}\right)$.