# How do I find the sum of the series -1^2-2^2+3^2+4^2-5^2… upto 4n terms? [duplicate]

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?

• Expanding is agood simple idea! Maybe your made an error in the expansion.
– user
May 7 '18 at 12:55

It seems that you should get to sum $32k+30$, not $8k+20$, and then simplify.

• I've obtained 32k+20! I check
– user
May 7 '18 at 13:00
• @gimusi That's right! :P May 7 '18 at 13:02

We have

$$(4k+3)^2+(4k+4)^2-(4k+1)^2-(4k+2)^2=16k^2+24k+9+16k^2+32k+16-16k^2-8k-1-16k^2-16k-4=32k+20$$

$(4k+3)^2-(4k+1)^2+(4k+4)^2-(4k+2)^2=(4k+3+4k+1)(2)+(4k+4+4k+2)(2)$

So the sum can be written as $2(1+2+3+\cdots+4n)$.

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 $$\sum_{k=0}^{n-1}(32k+20)= 32\frac{n(n-1)}{2}+20n=16n^2-12n$$

• Yes of course, I was considering from n=1!
– user
May 7 '18 at 13:05

In other terms,since $$(x+2)^2-x^2=4(x+1)$$ so you have to sum $$4(4k+2)+4(4k+3)=32k+20$$