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Use the binomial theorem to prove that $C(n,0) + C(n,1)\cdot 2 + C(n,2)\cdot 2^2 + \dots + C(n,n)\cdot 2^n = 3^n.$

I'm pretty confused with this concept, so any help would be greatly appreciated.

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  • $\begingroup$ $3^n= (2 + 1)^n$ $\endgroup$ – Mick Nov 11 '16 at 4:37
  • $\begingroup$ Hint: doesn't it look like $\binom{n}{0} + \binom{n}{1}x + \binom{n}{2} x^2 + \cdots$. $\endgroup$ – dxiv Nov 11 '16 at 4:38
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Binomial Theorem says that for $n\in\mathbb{N}$ $$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^{n}$$ for all $x\in\mathbb{R}$.

Put $x=2$.

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