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

Solution 1:

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: $$\sum_{k=1}^{4n-1}(-1)^{^{\frac {k(k+1)}2}}k=\overbrace{\sum_{k=1}^{4n}(-1)^{^{\frac {k(k+1)}2}}k}^{=4n}-4n=0$$

Solution 2:

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