Question

$\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}.....=$

• A. $\frac{2^n}{n+1}$
• B. $\frac{2^{n+1}-1}{n+1}$
• C. $\frac{2^{n+1}}{n+1}$
• D. None of these

Answer 1

The correct option is A $\frac{2^n}{n+1}$

$\begin{array}{c}\frac{C_o}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}.....\\ 1+\frac{n\times \left(n-1\right)}{3!}+\frac{n\times \left(n-1\right)\left(n-2\right)\left(n-3\right)}{5}+......\\ \frac{1}{n+1}\left[\left(n+1\right)\frac{\left(n+1\right)\times n\times \left(n-1\right)}{3!}+\frac{n\left(n+1\right)\left(n-1\right)\left(n-2\right)}{5!}\right]\\ \frac{1}{n+1}\left[2^{n+1-1}\right]\\ \frac{2^n}{n+1}\end{array}$

Hence the option $A$ is the correct answer.