Find the sum of the series:

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

I do the sum and each time I get a different answers, none of them matching with the answer in my book.

Book's answer:-n(2n+1)

My latest answer: -3n(2n-1)

My latest working:

(a)$\displaystyle 1^2, 3^2, 5^2, ...(2n-1)^2, so the series can be represented by (2r+1)^2 $

starting from r=1, there is (n-1) terms and 1^2 have to be added separately.

(b) $\displaystyle 2^2,4^2,6^2,...(2n)^2, so the series can be represented by (2r)^2$

starting from r=1, there is n terms.

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(a) $\displaystyle 1+\sum\limits_{r=1}^{n-1} (2n-1)^2$

$\displaystyle =1+4\sum\limits_{r=1}^{n-1} r^2-4\sum\limits_{r=1}^{n-1} r+\sum\limits_{r=1}^{n-1}1$

$\displaystyle =4/3 n^3-4n^2+11/3 n $

(b) $\displaystyle \sum\limits_{r=1}^{n} (2r)^2

=4/3n^3-2n^2+2/3n $

Summation=(a)-(b)=-6n^2+3n

=-3n(2n-1)

Can you please narrow down the field where I should search the error?

I am exasperated looking through my rough handwriting.