How to solve this series: $\sum_{k=0}^{2n+1} (-1)^kk $ $$\sum_{k=0}^{2n+1} (-1)^kk $$
The answer given is $-n-1$. I have searched for how to do it, but I have problems simplifying the sum and solving it.  How do you go about solving this?

Another approach is to exploit the sum $$\sum_{k=0}^{2n+1}x^{k}=\frac{1-x^{2n+2}}{1-x}.
 $$ Taking the derivative we have $$\sum_{k=0}^{2n+1}kx^{k}=\sum_{k=1}^{2n+1}kx^{k}=x\frac{-\left(2n+2\right)x^{2n+1}\left(1-x\right)+1-x^{2n+2}}{\left(1-x\right)^{2}}
 $$ hence taking $x=-1
 $ we get $$\sum_{k=0}^{2n+1}k\left(-1\right)^{k}=\color{red}{-\left(n+1\right)}$$ as wanted.
