Question

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

Answer 1

$$\begin{array}{c}\sum _{r=0}^n{3}^r{}^n{C}_r=4^n\\ [3^0{.}^n{C}_r+3^1{}^n{C}_1+3^2{}^n{C}_2+..........+3^n{}^n{C}_n]\\ [1+3n+9.\frac{n(n-1)}{2!}+.......{.3}^n.1][\because (1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+.....x^2]\\ (1+3{)}^n\end{array}$$