Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$

Question

Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$

I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.

Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$

Still couldn't get to the given answer of $-n(2n+1)$

Start wearing purple · Accepted Answer · 2013-08-24T11:35:29.700

8

Note that $$(2n-1)^2-(2n)^2=1-4n,$$ and therefore your sum is equal to $$\underbrace{(1+1+\ldots+1)}_{n\;\mathrm{times}}-4\left(1+2+\ldots+n\right)=n-4\cdot\frac{n(n+1)}{2}=-n(2n+1).$$ The only thing one needs to know is the arithmetic progression sum.

edited Aug 24 '13 at 11:35

answered Aug 24 '13 at 11:22

Start wearing purple

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where does the $n$ at the start of the second line come from please? – Gez Bishop Aug 24 '13 at 11:32
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@GezBishop It comes from $1$ in $1-4n$, which appears $n$ times. I have edited the answer to explain this. – Start wearing purple Aug 24 '13 at 11:34
I see, is that the same as in $\sum_{r=1}^n1=1+1+1+...+1=n$? thanks – Gez Bishop Aug 24 '13 at 11:39

score 2 · Answer 2 · answered Aug 24 '13 at 11:23

2

Hint 1: We have $$\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1).$$

Hint 2: Note that $$1^2-2^2+3^2-4^2+\cdots-(2n)^2=1^2+2^2+\cdots+(2n)^2-2(2^2+4^2+\cdots+(2n)^2)$$ $$ =1^2+2^2+\cdots+n^2-2^3(1^2+2^2+\cdots+n^2) =\sum_{k=1}^{2n}k^2-8\sum_{k=1}^{n}k^2.$$

answered Aug 24 '13 at 11:23

Paul

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In Hint 2, how do you get from the LHS to the RHS, please? – Gez Bishop Aug 24 '13 at 11:46
Wouldn't $n_1=1^2-2(2(1))^2=-4,n_2=2^2-2(2(2))^2=-48,...$ on that RHS? – Gez Bishop Aug 24 '13 at 12:00
@Gez What are you talking about? Do you disagree with $a-b+c-d=(a+b+c+d)-2(b+d)$ for example? – anon Aug 24 '13 at 14:33

score 1 · Answer 3 · answered Aug 24 '13 at 11:21

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Hint: $$1^2 - 2^2 + \ldots + (2n-1)^2 - (2n)^2 = 1^2 + 2^2 + \ldots + (2n)^2 - 8 (1^2 + 2^2 + \ldots + n^2)$$

answered Aug 24 '13 at 11:21

njguliyev

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score 1 · Answer 4 · answered Aug 24 '13 at 14:01

$${ 1 }^{ 2 }-{ 2 }^{ 2 }=-\left( 1+2 \right) \\ { 3 }^{ 2 }-{ 4 }^{ 2 }=-\left( 3+4 \right) \\ { 5 }^{ 2 }-{ 6 }^{ 2 }=-\left( 5+6 \right) \\ \vdots \\ { \left( 2n-1 \right) }^{ 2 }-{ \left( 2n \right) }^{ 2 }=-\left( 2n-1+2n \right) $$

Sum up, you get the answer that O.L. has written.

Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$

4 Answers4

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