Question

If $a_n=\sum _{n=0}^{\infty }\frac{1}{{}^{n}C_r}$then $\sum _{n=0}^{\infty }\frac{r}{{}^{n}C_r}$equals:

• A. $\left(n-1\right)a_n$
• B. $n{a}_n$
• C. $\frac{n}{2}{a}_n$
• D. none of these

Answer 1

The correct option is C

$\frac{n}{2}{a}_n$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $a_n=\sum _{n=0}^{\infty }\frac{1}{{}^{n}C_r}$

Let, $x=\sum _{n=0}^{\infty }\frac{r}{{}^{n}C_r}$

$\Rightarrow$ $x=\sum _{n=0}^{\infty }\frac{n-\left(n-r\right)}{{}^{n}C_r}$

$\Rightarrow$ $x=n\sum _{n=0}^{\infty }\frac{1}{{}^{n}C_r}-\sum _{n=0}^{\infty }\frac{n-r}{{}^{n}C_r}$

Step-2: Using formula to find the value:

$\Rightarrow$ $x=n\sum _{n=0}^{\infty }\frac{1}{{}^{n}C_r}-\sum _{n=0}^{\infty }\frac{n-r}{{}^{n}C_{n-r}}$

$\Rightarrow$ $x=n{a}_n-x$ ${}^{\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}\mathbf{-}\mathbf{r}}$

$\Rightarrow$ $2x=n{a}_n$

$\Rightarrow$ $x=\frac{n}{2}{a}_n$

$\Rightarrow$$\sum _{n=0}^{\infty }\frac{r}{{}^{n}C_r}=\frac{n}{2}{a}_n$

Hence, correct answer is option $\left(C\right).$