Question

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

Answer 1

$S=(n^2-1^2)+2(n^2-2^2)+3(n^2-3^2)+\dots \dots$

$S=n^2(1+2+3+\dots \dots )-(1^3+2^3+3^3+\dots \dots )=n^2\sum n-\sum n^3$

$S=n^2\cdot \frac{n(n+1)}{2}-{\left[\frac{n(n+1)}{2}\right]}^2$

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