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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,

$$ 2+2\sum_{k=1}^{(n-1)/2}{n \choose k} = 2^n $$


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That title..... – dot dot Jul 31 '12 at 19:30
Should be at least legible 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
up vote 3 down vote accepted

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 $n= 2m+1$, $$2\sum_{k=1}^{m} k \binom{2m+1}{k} = 2(2m+1) \sum_{k=0}^{m-1} \binom{2m}{k} = (2m+1) \left(\sum_{k=0}^{2m} \binom{2m}{k} - \binom{2m}{m} \right) = n \left(2^{n-1} - \binom{n-1}{(n-1)/2} \right)$$

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Thank you very much! – Adam Jul 31 '12 at 20:07
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}

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I think this was the equation that he already figured out, not the one that we were supposed to find. Also, some errors remain. – Rijul Saini Jul 31 '12 at 20:02
fixed the errors in his derivation (once it goes through peer review) – Adam Jul 31 '12 at 21:14

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