# Math Help - Combinatorics problem

1. ## Combinatorics problem

Hi guys, I'm new to this forum.

So I'm trying to solve this combinatorics/probability problem and I have arrived in a final form of:

[sigma(from i=0 to i=n) of nCi*2^i]/4^n

which is apparently equal to (3/4)^n

Does anyone of you know how sigma(from i=0 to i=n) of nCi*2^i = 3^n?

Thank you for your kind help.

2. Using the binomial theorem we get $\left( {2 + 1} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}2^k }$