Edit: To answer your question: No, the above is the summation up to 2n terms within the original series, it would not be changed at this point, but you would have to show your work I bet.
ie.
Spoiler:
The sum of squares, for 1 to n, is well known to be (1/6)n(n+1)(2n+1) so the sum of squares from 1 to 2n is (1/6)(2n)(2n+1)(4n+1)= (1/3)n(2n+1)(4n+1).
Here, however, we have a negative sign on all even squares so we have to subtract off twice.
But . Using the formula for sum of squares as before, that is 4(1/6)n(n+1)(2n+1)= (2/3)n(n+1)(2n+1). Notice that this is 1 to n, not 2n, now. Since we have to subtract that off twice, the original sum is
(1/3)n(2n+1)(4n+1)- (4/3)n(n+1)(2n+1)= (1/3)n(2n+1)(4n+1- 4n-4)= (1/3)n(2n+1)(-3)= -n(2n+1).
Check: If n= 3, 1- 4+ 9- 16+ 25- 36= -21
-3(2(3)+1)= -3(7)= -21.