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?
 A: 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*}
A: You have:
$$
\begin{align}
\sum_{k=0}^{2n+1} (-1)^k k &= \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} (2k+1) \\
&= \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} 1 \\
&=  - \sum_{k=0}^{n} 1 \\
&= -(n+1)
\end{align}
$$
A: Group the terms together in pairs:
$$\sum_{k=0}^{2n+1} (-1)^kk  = \sum_{k=0}^{n} (2k - (2k+1))  = \sum_{k=0}^{n} (-1) = -(n+1)$$
A: 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}
A: 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.
