Question

$\frac{C_{0}}{1}+\frac{C_{2}}{3}+\frac{C_{4}}{5}+\frac{C_{6}}{7}+ ........ =$

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

Answer 1

Answer: (C)

Putting the value of $C_0, C_2, C_4.....$, we get
$= 1 +\frac{n\left(n+1\right)}{3.2!} +\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5.4!} +..... = \frac{1}{n+1}$
$\left[\left(n+1\right)+ \frac{\left(n+1\right)n\left(n-1\right)}{3!} + \frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5!}+.....\right]$
Put $n + 1 = N$
$= \frac{1}{N} \left[ N +\frac{N\left(N-1\right)\left(N-2\right)}{3!}+\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)\left(N-4\right)}{5!}+.....\right]$
$= \frac{1}{N} \left\{^{N}C_{1}+^{N}C_{3}+^{N}C_{5} + .....\right\}$
$= \frac{1}{N} \left\{2^{N-1}\right\} = \frac{2^{n}}{n+1}\quad\left\{\because N = n+1\right\}$