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What is a closed form expression for $$\sum_{k=n/2+1}^n k {n \choose k}$$ I have looked at Sum of Binomial Coefficients Times a Polynomial which is pretty close but still does not solve this case where the sum is partial. Some hints are also in Sum of combinations of n taken k where k is from n to (n/2)+1 but I don't see how to combine the two.

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    $\begingroup$ just rewrite $k {n\choose k }= n {n-1 \choose k-1}$ $\endgroup$ – Exodd May 18 at 12:51
  • $\begingroup$ Can you expand more? Not clear to me... $\endgroup$ – user439907 May 18 at 12:53
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I am not sure what you mean by $\frac{n}{2}$ as that is not necessarily an integer, but presumably you mean that it is the floor or ceiling. In either case, you can use the identity $$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$$ for integers $n\ge k\ge 1$ on each expression and use the fact that the sum of the elements of the $n^{\text{th}}$ row of Pascal's triangle is $2^n.$ (Though you'll have to divide it by half and add or subtract a term, using the right-left symmetry of Pascal's triangle $\binom{p}{q}=\binom{p}{p-q}$)

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You have $S = S_n - S_{\frac{n}{2}}$, to get each sum follow the hint of @Exodd and use the closed-form of binomial sum for all terms between $0$ and (either) $n$ or $\frac{n}{2}$.

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