Question

If $C_0,C_1,C_2,....,C_n$ are binomial coefficients of order $n$, then the value of $\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+....=$

• A. $\frac{2^n+1}{n+1}$
• 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 $\frac{2^n}{n+1}$

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+......C_n{x}^n$

Integrating, We have

$\frac{(1+x{)}^{n+1}}{n+1}=C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+.....C_n\frac{x^{n+1}}{n+1}$

Putting, $x=1$

$\frac{(2{)}^{n+1}}{n+1}=C_0+C_1\frac{1}{2}+C_2\frac{1}{3}+....$

Putting, $x=-1$

$\frac{(0{)}^{n+1}}{n+1}=-C_0+C_1\frac{1}{2}+C_2\frac{(-1)}{3}+....$

Adding above two equations we have,

$\frac{(2{)}^{n+1}}{n+1}=2(\frac{C_1}{2}+\frac{C_3}{4}.....)$

$⟹\frac{(2{)}^n}{n+1}=\frac{C_1}{2}+\frac{C_3}{4}.....$

Hence, option D is correct.