I want to prove below:

n is natural number.

$$\sum_{k=1}^n k \binom{2n}{n+k} =\frac{1}{2}(n+1) \binom{2n}{n+1}$$

Please tell me above proof.


closed as off-topic by Grigory M, user91500, Namaste, Davide Giraudo, Hakim May 25 '14 at 13:36

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  • $\begingroup$ @takoika I hope I didn't mess up your question. $\endgroup$ – Santosh Linkha May 25 '14 at 8:00
  • $\begingroup$ Wolframalpha solves (1/2)(n+1)(2n_C_(n+1)).I want to know its process. $\endgroup$ – takoika May 25 '14 at 8:01
  • $\begingroup$ without $k$ this is partial sum of rows of Pascal triangle, which doesn't have a closed-form expression. This one probably doesn't have one either. You can asymptotics though. $\endgroup$ – Alex May 25 '14 at 9:58

You can use $$ \sum_{k=0}^{2n}\binom{2n}{k}=(1+1)^{2n}=4^n $$ and $$ \binom{m}{j}=\binom{m}{m-j} $$ and $$ j\binom{m}{j}=m\binom{m-1}{j-1} $$ to simplify your expression.

For instance, combining them to \begin{align} 2k\binom{2n}{n+k}&=(n+k)\binom{2n}{n+k}-(n-k)\binom{2n}{n-k}\\[1em] &=2n\binom{2n-1}{n+k-1}-2n\binom{2n-1}{n-k-1}\\[1em] &=2n\binom{2n-1}{n+k-1}-2n\binom{2n-1}{n+k} \end{align} Perhaps this can be used in a telescoping sum.

  • $\begingroup$ Is first fomula wrong? $\endgroup$ – takoika May 25 '14 at 14:18
  • $\begingroup$ Sorry, but yes. Shameful mistake. $\endgroup$ – LutzL May 25 '14 at 14:48

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