Please help me to solve this problem
Find the sum of the series up to 2n terms
1^2-2^2+3^2-4^2+5^2-7^2+.....
Here's one way it could be solved:
$\displaystyle \sum_{x=1}^{n}(2 x-1)^2 = \frac{1}{3} (4 n^3-n)$
$\displaystyle \sum_{x=1}^{n}(2 x)^2= \frac{2}{3} n (n+1) (2 n+1)$
$\displaystyle \sum_{x=1}^{n}(2 x-1)^2-\sum_{x=1}^{n}(2 x)^2 =\frac{1}{3} (4 n^3-n)-\frac{2}{3} n (n+1) (2 n+1)= -2 n^2-n$
That gives you 2 terms for each n, or 2n terms. I'm not certain this is how it's *supposed* to be solved, though
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. $\displaystyle \sum_{x=1}^{n}(2x)^2=4\cdot\sum_{x=1}^{n}x^2= 4\cdot\frac{n(n+1)(2n+1)}{6}=\frac{2}{3}n(n+1)(2n+ 1)$
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 $\displaystyle 4+ 16+ ...+ (2n)^2$ twice.
But $\displaystyle 4+ 16+ ...+ (2n)^2= 4(1)+ 4(4)+ ... 4(n^2)= 4(1+ 4+ ...+ n^2)$. 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.