How to solve this series: $\sum_{k=0}^{2n+1} (-1)^kk $

Group the terms together in pairs: $$\sum_{k=0}^{2n+1} (-1)^kk = \sum_{k=0}^{n} (2k - (2k+1)) = \sum_{k=0}^{n} (-1) = -(n+1)$$

Just write it out: \begin{aligned} \sum_{k=0}^{2n+1} (-1)^kk&=0-1+2-3+\cdots+2n-(2n+1)\\ &=(0-1)+(2-3)+\cdots+[2n-(2n+1)]\\ &=(-1)+(-1)+\cdots+(-1)\\ &=(n+1)(-1)\\ &=-n-1. \end{aligned}

Split it into two series:

\begin{align*} \sum\limits_{k=0}^{2n+1}(-1)^kk &=\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}2k+1\right)\\ &=\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}1\right)\\ &=-\sum\limits_{k=0}^{n}1\\ &=-(n+1)\,. \end{align*}