<s>[INST] How do I find the sum of the series -1^2-2^2+3^2+4^2-5^2… upto 4n terms? I tried by giving 
$$ S = \sum_{k=0}^{n-1} \left((4k+3)^2+(4k+4)^2-(4k+1)^2-(4k+2)^2\right)  $$
but I am stuck here. I have no idea what to do next. The answer in my book says 4n(n+1). How can I get it? I tried expanding (4k+1)^2, etc. and got $ \sum_{k=0}^{n-1} (8k+20) $, I tried to further expand this by taking it as $ 8\sum_{k=0}^{n-1}k + 20n = 8(n-1)(n)/2 + 20n = 4n^2+16n= 4n(n+4) $ which is not the right answer. What have I done wrong? What must I do now?
 [/INST] Expand
\begin{align}
&(4k+3)^2-(4k+2)^2+(4k+4)^2-(4k+1)^2\\
\qquad&=(4k+3-4k-2)(4k+3+4k+2)+(4k+4-4k-1)(4k+4+4k+1)\\
\qquad&=8k+5+3((8k+5)\\
\qquad&=4(8k+5)\\
\qquad&=32k+20
\end{align}
So your sum is
$$
\sum_{k=0}^{n-1}(32k+20)=
32\frac{n(n-1)}{2}+20n=16n^2-12n
$$
</s>