Question

Prove that
$1+\frac{2n}{3}+\frac{2n(2n+2)}{3.6}+\frac{2n(2n+2)(2n+4)}{\mathrm{3.6.9}}+.....$
$=2^n\{1+\frac{n}{3}+\frac{n(n+1)}{3.6}+\frac{n(n+1)(n+2)}{\mathrm{3.6.9}}+....\}$

Answer 1

given that

$RHS=2^n[1+\frac{n}{3}+\frac{n(n+1)}{3.6}+\frac{n(n+1)(n+2)}{\mathrm{3.6.9}}+.....]$

$=2^n[1+\frac{n}{3}+\frac{n(n+1)}{2!{.3}^2}+\frac{n(n+1)(n+2)}{3!{.3}^3}+.....]$

given expansion is of the form $(1+x{)}^{-n}$ where

$x=\frac{-1}{3}$

\therefore

$RHS=2^n[1-\frac{1}{3}{]}^{-n}$

$=2^n[\frac{2}{3}{]}^{-n}$

$=[\frac{1}{3}{]}^{-n}=[1-\frac{2}{3}{]}^{-n}$

let us expand it

$RHS=[1-\frac{2}{3}{]}^{-n}$

$=[1+\frac{2n}{3}+\frac{n(n+1).4}{2!{.3}^2}+\frac{n(n+1)(n+2).8}{3!{.3}^3}+.....]$

$=[1+\frac{2n}{3}+\frac{2n(2n+2)}{3.6}+\frac{2n(2n+2)(2n+4)}{\mathrm{3.6.9}}........]$

$=LHS$