There is probably something stupidly simple I'm missing, but I'm trying to find a closed form for:
$ 2\sum_{k=1}^{(n-1)/2} k \, {n \choose k} \hspace{1cm} (n\textrm{ is odd}) $
Anyone know how to do this?
I've figured out that since $n$ is odd,
$ 2+2\sum_{k=1}^{(n-1)/2}{n \choose k} = 2^n $
Thanks...