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Math Help - Combinatorics problem

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