# What is sum nC0+2(nC1)+----+2^n(nCn)=?

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`(1+x)^n=^nC_0 1^nx^0+^nC_1 1^(n-1)x^1+...+^nC_n 1^0 x^n`

`=sum_{r=0}^{r=n } (^nC_r) 1^(n-r)x^r` (i)

Thus if we substitute x=2 in (i) ,then we have

`sum_(r=0)^(r=n) {^nC_r }1^(n-r) 2^r=(1+2)^n`

`^nC_0 1^n+^nC_1 1^(n-1) 2^1+........+^nC_n 1^0 2^n=3^n`

`^nC_0+2 (^nC_1)+...+2^n ^nC_n=3^n`

`` Thus sum is `3^n`

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