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Find the sum $S=\sum_{k=1}^{2015}{(-1)^{\frac{k(k+1)}{2}}}k.$

I partitioned k in two categories Either k is congruent to 0 , 3 mod(4) or congruent to 1,2 mod(4). But still I didn't get answer

Martin Sleziak
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Tushar
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3 Answers3

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The sum is $\sum_{j=0}^{503}\sum_{k=4j}^{4j+3}(-1)^{k (k+1)/2}k$ (don't worry about $k=0$... it's equal to zero)

This is $\sum_{j=0}^{503}[4k-(4k+1)-(4k+2)+(4k+3)]=\sum_{j=0}^{503}0=0$.

fleablood
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Interesting question.

Each group of four terms sum to $4$. Hence the sum to $4n$ terms equals $4n$, e.g. $$\sum_{k=1}^{2016}(-1)^{^{\frac {k(k+1)}2}}k=2016$$ and it follows that

$$\sum_{k=1}^{2015}(-1)^{^{\frac {k(k+1)}2}}k=\sum_{k=1}^{2016}(-1)^{^{\frac {k(k+1)}2}}k-2016=\color{red}0$$

NB: It is interesting to note the general case as follows: $$\sum_{k=1}^{4n-1}(-1)^{^{\frac {k(k+1)}2}}k=\overbrace{\sum_{k=1}^{4n}(-1)^{^{\frac {k(k+1)}2}}k}^{=4n}-4n=0$$

Hypergeometricx
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  • Might be worthwhile noting that as $(-1)^{0(0+1)/2}*0 = 0$ that $\sum_{i=1}^{2015}(-1){k(k+1)/2}k = \sum_{i=0}^{2015}(-1){k(k+1)/2}k$ and so of you start and mod 0 rather than mod 4 then i) each group adds to 0 instead of 4, ii) your groups are complete, not incomplete. – fleablood May 25 '17 at 06:39
  • @fleablood Yes, thanks, that's a good way of looking at it too. – Hypergeometricx May 25 '17 at 06:48
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The series is $-1-2+3+4-5-6+7+8-\cdots$. A quartet of four terms $$-(4k-3)-(4k-2)+(4k-1)+(4k)$$ contributes $4$ to the sum, etc.

Angina Seng
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  • yes , but given answer is 0. – Tushar May 25 '17 at 05:39
  • (0-1-2+3) + (4-5-6+7)+......+(2012-2013-2014+2015) = 0. (-1-2+3+4)+(-5-6+7+8) + ...... +(-2009-2010+2011+2012) + (-2013-2014+2015 + 2016) = 4504 = 2016. But the sum ends at 2015 so (-1-2+3+4)+(-5-6+7+8) + ...... +(-2009-2010+2011+2012) + (-2013-2014+2015) = 4503 -2013-2014+2015 = 2012 -2013 -2014 + 2015 = 0. – fleablood May 25 '17 at 06:30
  • @fleablood You said that in your answer! – Angina Seng May 25 '17 at 06:31
  • Or shorter. The number and four each quartet. But the last quartet is incomplete. So doing it by quartets starting at 1 is 4*503 + (4 - 2016) = 0. If you start at 0, however the quartets add to 0. ANd the quartets are all complete. – fleablood May 25 '17 at 06:32
  • No I didn't. I am comparing the two methods and explaining how one answer can add 4 each time and still come up with 0 by not completing. I didn't mention that in my answer. – fleablood May 25 '17 at 06:35