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.