Question

The value of $\frac{{{}^{n}C}_1}{2}+\frac{{{}^{n}C}_3}{4}+\frac{{{}^{n}C}_5}{6}+...$ is

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

Answer 1

Answer: (C)

$\frac{2^n-1}{n+1}$

The $rth$ term in the given expression is

$T_r=\frac{{{}^{n}C}_{2r-1}}{2r}$

Since $\frac{1}{r+1}.{{}^{n}C}_r=\frac{1}{n+1}.{{}^{n+1}C}_{r+1}$

$\therefore T_r=\frac{{{}^{n}C}_{2r-1}}{2r}=\frac{1}{n+1}.{{}^{n+1}C}_{2r}$

$\therefore \frac{{{}^{n}C}_1}{2}+\frac{{{}^{n}C}_3}{4}+\frac{{{}^{n}C}_5}{6}+....=\frac{1}{n+1}[{{}^{n+1}C}_2+{{}^{n+1}C}_4+...]$

$=\frac{1}{n+1}[2^{n+1-1}-{{}^{n}C}_0]=\frac{2^n-1}{n+1}$