Question

Value of the expression $\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+...$ is

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

Answer 1

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

$(1+x{)}^n=1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{}^n{C}_n(x^n)$
$(1-x{)}^n=1-{}^n{C}_{1}x+{}^n{C}_2{x}^2+...(-1{)}^n{}^n{C}_n\left(x^n\right)$
Therefore
$\frac{1}{2}\left[\right(1+x{)}^n-(1-x{)}^n]={}^n{C}_{1}x+{}^n{C}_3{x}^3+...$
Integrating with respect to x, we get
$\frac{1}{2(n+1)}\left[\right(1+x{)}^{n+1}+(1-x{)}^{n+1}]{|}_0^1=\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_3}{4}+\frac{{}^n{C}_5}{6}...$
Hence
$\frac{1}{2(n+1)}\left[\right(1+x{)}^{n+1}+(1-x{)}^{n+1}]{|}_0^1$
$=\frac{1}{2(n+1)}[2^{n+1}+0-1-1]$
$=\frac{1}{n+1}[2^n-1]$