Can we sum $1^2-2^2+3^2-4^2+\dotsb+ (2n-1)^2-(2n)^2$?

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