Question

$\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+...$ is equal to

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

Answer 1

Answer: (C)

We know, ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...$
and ${(1-x)}^n=C_0-C_{1}x+C_2{x}^2+...$
On adding, we get ${(1+x)}^n+{(1-x)}^n=2[C_0+C_2{x}^2+...]$
On integrating $0$ to $1$ , we get
${\left[\frac{{(1+x)}^{n+1}}{n+1}-\frac{{(1-x)}^{n+1}}{n+1}\right]}_0^1$
$=2{\left[\frac{C_{0}x}{1}+\frac{C_2{x}^3}{3}+...\right]}_0^1$
$\Rightarrow$ $\frac{2^{n+1}}{n+1}=2\left[\frac{C_0}{1}+\frac{C_2}{3}+...\right]$
$\Rightarrow$ $\frac{C_0}{1}+\frac{C_2}{3}+...=\frac{2^n}{n+1}$