Question

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

Answer 1

$\sum _{r=0}^n{}^n{C}_r{3}^r={}^n{C}_0{3}^0+{}^n{C}_1{3}^1+{}^n{C}_2{3}^2+.....+{}^n{C}_n{3}^n=(1+3{)}^n=4^n$