Question

The value of $\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\cdots$ equals to

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

Answer 1

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

We know that
$\frac{(1+x{)}^n-(1-x{)}^n}{2}=C_{1}x+C_3{x}^3+C_5{x}^5+\cdots$
Integrating from x = 0 to x = 1, we get
$\frac{1}{2}{\int }_0^1\left\{\right(1+x{)}^n-(1-x{)}^n\}dx$
${\int }_0^1(C_{1}x+C_3{x}^3+C_5{x}^5+\cdots )dx$
$\Rightarrow \frac{1}{2}{\{\frac{(1+x{)}^{n+1}}{n+1}+\frac{(1-x{)}^{n+1}}{n+1}\}}_0^1=\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\cdots$
$or\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\cdots =\frac{1}{2}\{\frac{2^{n+1}}{n+1}+\frac{0-1}{n+1}\}$
$=\frac{1}{2}\left(\frac{2^{n+1}-2}{n+1}\right)=\frac{2^n-1}{n+1}$