# modified $\sum{k{n \choose k}}$ closed form expression

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...

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That title..... – dot dot Jul 31 '12 at 19:30
Should be at least legible now...is there something else wrong with it? – Adam Jul 31 '12 at 19:31
Note that $k \binom{n}{k} = n \binom{n-1}{k-1}$, and proceed. – Rijul Saini Jul 31 '12 at 19:31
Don't worry, now's fine – dot dot Jul 31 '12 at 19:32

If you want a complete solution, using the equation in the above comments i.e. $k \binom{n}{k} = n \binom{n-1}{k-1}$, we get that if $n= 2m+1$, $$2\sum_{k=1}^{m} k \binom{2m+1}{k} = 2(2m+1) \sum_{k=0}^{m-1} \binom{2m}{k} = (2m+1) \left(\sum_{k=0}^{2m} \binom{2m}{k} - \binom{2m}{m} \right) = n \left(2^{n-1} - \binom{n-1}{(n-1)/2} \right)$$
I think the last expression can be simplified as $n{n-1\choose m}$. – Tyler Aug 8 '12 at 16:13
If $n$ is odd we have (where $j=\frac{n-1}{2}$): \begin{eqnarray} 2^n &=& \binom{n}{0}+ \binom{n}{1} +\binom{n}{2} + \cdots +\binom{n}{n-2} +\binom{n}{n-1}+\binom{n}{n} \\ &=& \left \{\binom{n}{0} + \binom{n}{n} \right \} +\left[ \left \{ \binom{n}{1} + \binom{n}{n-1}\right \} + \cdots + \left \{\binom{n}{j} + \binom{n}{j+1} \right \} \right]\\ &=& 2+ 2\sum_{k=1}^{(n-1)/2} \binom{n}{k}. \end{eqnarray}