Prove that $sum_{r=0}^{n} 3^{r ~}{ }^{n} C_{r}=4^{n}$.


Prove that $\sum_{r=0}^{n} 3^{r ~}{ }^{n} C_{r}=4^{n}$.


By Binomial Theorem,

$\sum_{r=0}^{n}{ }^{n} C_{r} a^{n-r} b^{r}=(a+b)^{n}$

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

$\sum_{r=0}^{n}{ }^{n} C_{r}(1)^{n-r}(3)^{r}=(1+3)^{n}$

$\Rightarrow \sum_{r=0}^{n} 3^{r^{n}} C_{r}=4^{n}$

Hence, proved.

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