Question

$\underset{r=0}{\sum ^n}\left(\frac{r+2}{r+1}\right){.}^n{C}_r$ is equal to :

• A. $\frac{2^n(n+2)-1}{(n+1)}$
• B. $\frac{2^n(n+1)-1}{(n+1)}$
• C. $\frac{2^n(n+3)-1}{(n+1)}$
• D. $0$

Answer 1

The correct option is C $\frac{2^n(n+3)-1}{(n+1)}$

$I=\sum _{r=0}^n(\frac{r+2}{r+1}{)}^n{C}_r$

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

$I\Rightarrow \sum _{r=0}^n{}^n{C}_r+\sum {}_{r=0}^n\frac{{}^n{C}_r}{(r+1)}$

$(1+x{)}^n{=}^n{C}_0(1{)}^n{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{.}^n{C}_n{x}^n$

Put $x=1$

$\Rightarrow (1+1{)}^n{=}^n{C}_0{+}^n{C}_1{+}^n{C}_2+.....{.}^n{C}_n.$

$2^n=\sum _{r=0}^n{}^n{C}_r$

$\Rightarrow A=2^x$

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+....{.}^n{C}_n{x}^n$

integrate both side

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

Put $x=0$

$\frac{(1{)}^{n+1}-1}{(n+1)}=P$

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

Put $x=1$

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

$\Rightarrow \sum {}_{r=0}^n\times \frac{1}{(r+1)}\times {}^n{C}_r=\frac{2^{(n+1)-1}}{(n+1)}\Rightarrow B=\frac{2^{(n+1)}-1}{(n+1)}$

$I=A+B=2^n+\frac{2^{(n+1)}-1}{(n+1)}$

$I=\frac{2^n(n+1)+2^{n+1}-1}{n+1}$

$I=\frac{2^n(n+3)-1}{(n+1)}$