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Write $\sum\limits_{k=\frac{n+1}{2}}^n{n \choose k}$ in its closed form.
$n \in N_{odd}$

First time to confront this kind of problem. How do I solve it?

(If its a "you are asking for too much" question, I will much appreciate a link to a good explanatory site on the subject.)

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    @MichaelS:Alright,then for$n$is odd,simple induction shows the closed form to be $2^{(n-1)}$.2019-03-04

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Writing $2n+1$ instead of $n$ gives $\sum\limits_{k=n+1}^{2n+1}{2n+1 \choose k}$ and that is exactly a half of the summands in the expansion of $(1+1)^{2n+1}$, so the answer is $4^n$. Or, in the initial terms, $2^{n-1}$.