Suma de combinaciones de n tomadas k donde k es de n a (n/2)+1

Recordar que

$$\sum_{k=0}^n \dbinom{n}k = 2^n$$ Además, recuerda que $$\dbinom{n}k = \dbinom{n}{n-k}$$ Por lo tanto, para impares $n$, tenemos $$\begin{align} 2^n & = \sum_{k=0}^n \dbinom{n}k\\ & = \sum_{k=0}^{(n-1)/2} \dbinom{n}k + \sum_{k=(n+1)/2}^n \dbinom{n}k\\ & = \sum_{k=0}^{(n-1)/2} \dbinom{n}{n-k} + \sum_{k=(n+1)/2}^n \dbinom{n}k\\ & = \sum_{k=(n+1)/2}^n \dbinom{n}k + \sum_{k=(n+1)/2}^n \dbinom{n}k\\ & = 2\sum_{k=(n+1)/2}^n \dbinom{n}k \end{align}$$ Por lo tanto, si $n$es raro, tenemos $$\sum_{k=(n+1)/2}^n \dbinom{n}k = 2^{n-1}$$ Si $n$es par, tenemos $$\begin{align} 2^n & = \sum_{k=0}^n \dbinom{n}k\\ & = \sum_{k=0}^{n/2-1} \dbinom{n}k + \dbinom{n}{n/2} + \sum_{k=n/2+1}^n \dbinom{n}k\\ & = \sum_{k=0}^{n/2-1} \dbinom{n}{n-k} + \dbinom{n}{n/2} + \sum_{k=n/2+1}^n \dbinom{n}k\\ & = \sum_{k=n/2+1}^n \dbinom{n}k + \sum_{k=n/2+1}^n \dbinom{n}k + \dbinom{n}{n/2}\\ & = 2\sum_{k=n/2+1}^n \dbinom{n}k + \dbinom{n}{n/2} \end{align}$$ Por lo tanto, si $n$es par, tenemos $$\sum_{k=n/2+1}^n \dbinom{n}k = 2^{n-1} - \dfrac12 \dbinom{n}{n/2}$$