Prove that $\sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $

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By Binomial Theorem,

$\sum\limits_{r = 0}^n {{\,^n}{C_r}{a^{n - r}}{b^r} = {{\left( {a + b} \right)}^n}} $

By putting $b=3$ and $a=1$ in the above equation, we obtain

$\sum\limits_{r = 0}^n {{\,^n}{C_r}{{\left( 1 \right)}^{n - r}}{{\left( 3 \right)}^r} = {{\left( {1 + 3} \right)}^n}} $

$ \Rightarrow \sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $

Hence proved.

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