Question

If $(1+x{)}^n={\sum }_{r=0}^n{C}_r{x}^r$
Prove that $\frac{2^2{C}_0}{1.2}+\frac{2^3{C}_1}{2.3}+......+\frac{2^{n+2}{C}_n}{(n+1)(n+2)}$ =?

Answer 1

${(1+x)}^n={\sum }_{r=0}^n{c}_r{x}^r$

$\Rightarrow \int {(1+x)}^n={\sum }_{r=0}^n{c}_r{x}^r$ [ By integrating both side ]

$\Rightarrow \frac{{(1+x)}^{n+1}}{n+1}={\sum }_{r=0}^n\frac{c_r{x}^{r+1}}{r+1}$

$\Rightarrow \frac{1}{n+1}\int {(1+x)}^{n+1}={\sum }_{r=0}^n\frac{c_r}{r+1}\int x^{r+1}$ [Again integration ]

$\Rightarrow \frac{1}{(n+1)(n+2)}{(1+x)}^{n+2}={\sum }_{r=0}^n\frac{c_r}{(r+1)(r+1)}.x^{r+2}$

$\Rightarrow \frac{C_0}{1.2}{x}^2+\frac{C_1}{2.3}.x^3+......+\frac{C_n}{(n+1)(n+2)}.x^{n+2}=\frac{{(1+x)}^{n+2}}{(n+1)(n+2)}$

by putting $x=2$ the equation become

$\Rightarrow \frac{C_0}{1.2}{2}^2+\frac{C_1}{2.3}.2^3+......+\frac{C_n}{(n+1)(n+2)}.2^{n+2}=\frac{3^{n+2}}{(n+1)(n+2)}$

$\therefore \frac{3^{n+2}}{(n+1)(n+2)}$

Hence, proved.