Question

If ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$ then find the value of $\frac{1}{2}{C}_1+\frac{1}{4}{C}_3+\frac{1}{6}{C}_5+....$

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

Answer 1

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

Given that ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$
Now, $\frac{{(1+x)}^n-{(1-x)}^n}{2}=C_{1}x+C_3{x}^3+C_5{x}^5+...$
Integrating wrt x from $0$ to $1$
we get ${\int }_0^1\left(\frac{{(1+x)}^n-{(1-x)}^n}{2}\right)dx=\frac{2^{n+1}-1-(0-1)}{2(n+1)}=\frac{1}{2}{C}_1+\frac{1}{4}{C}_3+\frac{1}{6}{C}_5+....$
$\Rightarrow \frac{1}{2}{C}_1+\frac{1}{4}{C}_3+\frac{1}{6}{C}_5+....=\frac{2^n}{n+1}$