i am trying to prove this statement for all $n \in \mathbb{N}$ with the help of induction:
$4 \sum_{k=1}^{n} (-1)^kk=(-1)^n(2n+1)-1$
base case: n=1
$4 \sum_{k=1}^{1} (-1)^11=-4=(-1)^1(2*1+1)-1$ .. OK
induction hypothesis: for all $n \in \mathbb{N}$ let be $4 \sum_{k=1}^{n} (-1)^kk=(-1)^n(2n+1)-1$
inductive step: $n \rightarrow n+1$
$4 \sum_{k=1}^{n+1} (-1)^kk=4 \sum_{k=1}^{n} (-1)^kk+(-1)^{n+1}(n+1)=(-1)^n(2n+1)-1+(-1)^{n+1}(n+1)=...help...=(-1)^{n+1}(2(n+1)+1)-1$
i need help for $..help..$
thanks a lot