MathDB

2017 Kosovo TST Problem 4

Source:

March 19, 2017

combinatorics

Problem Statement

For every

n \in \mathbb{N}_{0}

, prove that

\sum_{k=0}^{\left[\frac{n}{2} \right]}{2}^{n-2k} \binom{n}{2k}=\frac{3^{n}+1}{2}