# Q14.    Prove that $\sum_{r = 0}^n3^r \ ^nC_r = 4^n$

N neha

As we know from Binomial Theorem,

$\sum_{r = 0}^na^r \ ^nC_r = (1+a)^n$

Here putting a = 3, we get,

$\sum_{r = 0}^n3^r \ ^nC_r = (1+3)^n$

$\sum_{r = 0}^n3^r \ ^nC_r = 4^n$

Hence Proved.

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