Question

If $\sum _{r=1}^n{}^n{C}_r{3}^r$ is equal to $4095$ then $n$ equals.

Answer 1

We have,

${}_{r=1}^n{\sum }^n{C}_r=4095$

$⟹{}^n{C}_1{+}^n{C}_2{3}^2+...{.}^n{C}_n{3}^n=4095$

$⟹(1+3{)}^n{-}^n{C}_0=4095$

$⟹4^n-1=4095$

$⟹4^n=4^6$

$⟹n=6$

Hence, this is the answer.