How to do this?
$\displaystyle \sum_{k=1}^{100}(-1)^{k}k$
To make a little clearer, just like bandekrait did, note that $\displaystyle \sum_{\ell=1}^{100}(-1)^{\ell}\ell=\sum_{\ell=1}^{50}(-1)^{2\ell+1}(2\ell-1)+\sum_{\ell=1}^{50}(-1)^{2\ell}2\ell$ (the evens in the odds) but $\displaystyle (-1)^{2\ell-1}=-1,(-1)^{2\ell}=1$ so this becomes $\displaystyle \sum_{\ell=1}^{50}-\left\{2\ell-1\right\}+\sum_{\ell=1}^{50}2\ell=\sum_{\ell=1}^{5 0}1=50$.