Review question

# Can we sum $1^2-2^2+3^2-4^2+\dotsb+ (2n-1)^2-(2n)^2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9819

## Question

Prove that the sum of the first $n$ terms of the arithmetical progression $a+(a+d)+(a+2d)+...,$ is $\frac{1}{2}n\lbrace{2a+(n-1)d}\rbrace.$

Hence, or otherwise, show that $1^2-2^2+3^2-4^2+\dotsb+ (2n-1)^2-(2n)^2=-n(2n+1).$

Deduce the sums of the series

1. $1^2-2^2+3^3-4^2+\dotsb+(2n-1)^2-(2n)^2+(2n+1)^2$,

and

2. $21^2-22^2+23^2-24^2+\dotsb+39^2-40^2$.