Question

Prove that $\Rightarrow {}^n{C}_0+\frac{3{}^n{C}_1}{2}+\frac{3^2{}^n{C}_2}{3}...=\frac{(1+3{)}^{n+1}-1}{3(n+1)}=\frac{4^{n+1}-1}{3(n+1)}$

Answer 1

$(1+3x{)}^n={}^n{C}_0+{}^n{C}_1\left(3x\right)+{}^n{C}_2(3x{)}^2+........+{}^n{C}_n(3x{)}^n----(1)$

Intergrating eq (1) w.r To $x$ with limit $0$ to $1$ we get

${\left[\frac{(1+3x{)}^{n+1}}{3(n+1)}\right]}_0^1={}^n{C}_{0}x+\frac{3\cdot {}^n{C}_1(x{)}^2}{2}+\frac{3^2\cdot {}^n{C}_2(x{)}^3}{3}+........+\frac{3^n\cdot {}^n{C}_n(x{)}^{n+1}}{n+1}----\left(2\right)$

$\frac{(1+3{)}^{n+1}-1}{3(n+1)}={}^n{C}_{0}x+\frac{3\cdot {}^n{C}_1(x{)}^2}{2}+\frac{3^2\cdot {}^n{C}_2(x{)}^3}{3}+........+\frac{3^n\cdot {}^n{C}_n(x{)}^{n+1}}{n+1}$

Hence Proved