# Partial sum of binomial coefficients n choose k multplied by k for k from n/2+1 to n

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.

• just rewrite $k {n\choose k }= n {n-1 \choose k-1}$ – Exodd May 18 at 12:51
• Can you expand more? Not clear to me... – user439907 May 18 at 12:53

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 $$\binom{p}{q}=\binom{p}{p-q}$$)
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 $$\frac{n}{2}$$.