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This question already has an answer here:

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?

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marked as duplicate by Jyrki Lahtonen May 7 '18 at 13:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Expanding is agood simple idea! Maybe your made an error in the expansion. $\endgroup$ – user May 7 '18 at 12:55
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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 $$ \sum_{k=0}^{n-1}(32k+20)= 32\frac{n(n-1)}{2}+20n=16n^2-12n $$

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  • $\begingroup$ Yes of course, I was considering from n=1! $\endgroup$ – user May 7 '18 at 13:05
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It seems that you should get to sum $32k+30$, not $8k+20$, and then simplify.

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  • $\begingroup$ I've obtained 32k+20! I check $\endgroup$ – user May 7 '18 at 13:00
  • $\begingroup$ @gimusi That's right! :P $\endgroup$ – Macrophage May 7 '18 at 13:02
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We have

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

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

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

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

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