Question

Find the sum of the first $2n$ terms of the series $1^2-2^2+3^2-4^2+....$

Answer 1

Given,

$1^2-2^2+3^2-4^2+...............$

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

the series can be re-written as,

$(1+2)(1-2)+(3-4)(3+4)+...........+(n-1-n)(n-1+n)$

$-1(1+2+3+4+..........+(n-1)+n)$

$-1\times \frac{n(n+1)}{2}$

$=-\frac{n(n+1)}{2}$, when n is even.

$=\frac{n(n+1)}{2}$, when n is odd.