Closed form of $\sum_{k=1}^{2015}(-1)^{k(k+1)/2}k$

Question

$$\sum_{k=1}^{2015}(-1)^\frac{k(k+1)}{2}\times k$$

How to solve this. Answer provided is $0$

When is $\frac{k(k+1)}{2}$ even? When is it odd? Note that $\frac{k(k+1)}{2}=1+2+\cdots+k$. Try splitting the sum in the positive and negative terms. — , Feb 15 '16 at 09:31
See also: Find the sum of finite series $S=\sum_{k=1}^{2015}{(-1)^{\frac{k(k+1)}{2}}}k$. — Martin Sleziak, May 25 '17 at 08:57

score 3 · Accepted Answer · answered Feb 15 '16 at 12:23

3

$\overbrace{-1-2+3+4}^4\overbrace{-5-6+7+8}^4\dots\overbrace{-2013-2014+2015+2016}^4=2016$

Take off the last $2016$, and we get

$\overbrace{-1-2+3+4}^4\overbrace{-5-6+7+8}^4\dots-2013-2014+2015=0$

answered Feb 15 '16 at 12:23

robjohn

345,667

Alex · Answer 2 · 2016-02-15T09:48:24.577

2

You get $-1-2 +3 +4 -5-6+7+8$ etc, so you get essentially $4+4+ \ldots 4 \ \ 503 $ times + 3 more terms

edited Feb 15 '16 at 09:48

answered Feb 15 '16 at 09:36

Alex

19,262

score 1 · Answer 3 · answered Feb 15 '16 at 09:33

Note that if $$k=4n, n\in \mathbb N, (-1)^{\frac{k(k+1)}{2}}=(-1)^{2n(4n+1)}=1$$ $$k=4n+1, n\in \mathbb N, (-1)^{\frac{k(k+1)}{2}}=(-1)^{(2n+1)(4n+1)}=-1$$ $$k=4n+2, n\in \mathbb N, (-1)^{\frac{k(k+1)}{2}}=(-1)^{(4n+3)(2n+1)}=-1$$ $$k=4n+3, n\in \mathbb N, (-1)^{\frac{k(k+1)}{2}}=(-1)^{2(n+1)(4n+3)}=1$$

Thus, you can find the sum of four arithmetic progressions to solve the problem.

Closed form of $\sum_{k=1}^{2015}(-1)^{k(k+1)/2}k$

3 Answers3

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