More fun in 2016, Part 10

simple. write as $1+(-3+6)+(-10+15)+...+(-1953+2016)$ $=1+3+5+...+63$ we know that the sum of the first n odds is $n^2$ . so the summation is $=32^2=\boxed{1024}$

$\begin{aligned} S & = 1-3+6-10+\ldots-1953 \\ & = \displaystyle \sum^{62}_{n=1}(-1)^{n+1}\frac{n(n+1)}{2} \\ & = \frac{1}{2}\left(\displaystyle \sum^{62}_{n=1}(-1)^{n+1}n^2+ \displaystyle \sum^{62}_{n=1}(-1)^{n+1}n\right) \\ & = \frac{1}{2}\left( \displaystyle \sum^{31}_{n=1}(2n-1)^2- \displaystyle \sum^{31}_{n=1}(2n)^2+ \displaystyle \sum^{62}_{n=1}(-1)^{n+1}n\right) \\ & = \frac{1}{2}\left(4 \displaystyle \sum^{31}_{n=1}n^2-4 \displaystyle \sum^{31}_{n=1}n+ \displaystyle \sum^{31}_{n=1}1-4 \displaystyle \sum^{31}_{n=1}n^2 + \displaystyle \sum^{62}_{n=1}(-1)^{n+1}n\right) \\ & = \frac{1}{2}\left(-4 \displaystyle \sum^{31}_{n=1}n+ \displaystyle \sum^{31}_{n=1}1+ \underbrace{\underbrace{\displaystyle \sum^{62}_{n=1}(-1)^{n+1}n}_{\underbrace{1-2}_{-1}+\underbrace{3-4}_{-1}+\ldots+\underbrace{61-62}_{-1}}}_{31(-1)=-31}\right) \\ & = \frac{1}{2}\left(-4\cdot \frac{31×32}{2}+31-31\right) \\ & = -992 \\ \Rightarrow S+2016 & = \boxed{1024} \end{aligned}$

XPLORE: sum((-1)^(k+1)*sum(j , j=1 to k), k=1 to 63) quickly delivers 1024.