Question

Prove that $\underset{r=0}{\sum ^n}{3}^r{}^n{C}_r=4^n$

Answer 1

Consider the given expression,

$\sum _{r=0}^n{3}^r{}^n{C}_r=4^n$

Binomial theorem,

$\sum _{r=0}^n{}^n{C}_r{a}^{n-r}{b}^r={(a+b)}^n$

Put $a=3$ and $b=1$, we get

$\sum _{r=0}^n{}^n{C}_r{1}^{n-r}{3}^r={(1+3)}^n$

$\sum _{r=0}^n{}^n{C}_r{3}^r={\left(4\right)}^n$

$\sum _{r=0}^n{3^r}^n{C}_r={\left(4\right)}^n$

Hence, proved