Question

Write the sum of the series : $1^2-2^2+3^2-4^2+5^2-6^2+...+(2n-1{)}^2-(2n{)}^2$

Answer 1

$1^2-2^2+3^2-4^2+5^2-6^2+...+(2n-1{)}^2-(2n{)}^2$

We have, $a^2-b^2=(a-b)(a+b)$

$1^2-2^2+3^2-4^2+5^2-6^2+...+(2n-1{)}^2-(2n{)}^2$

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

$=(-1)(1+2)+(-1)(3+4)+...(-1)(2n-1+2n)=-1[1+2+3+4+...2n-1+2n]$

$=-1\left[\frac{2n(2n+1)}{2}\right][\because 1+2+\mathrm{3.....}x=\sum x=\frac{x(x+1)}{2}]$

$=-n(2n+1)$

Hence, sum of the series $=-n(2n+1)$