Question

If $C_r$ means ${}^n{C}_r$ then $\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\cdots =$

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

Answer 1

The correct option is C

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

Consider the expansion
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+\cdots +C_n{x}^n$ . . .(i)
Integrating both sides of (i) within limits - 1 to 1, we get
${\int }_{-1}^1(1+x{)}^{n}dx={\int }_{-1}^1(C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+\cdots +C_n{x}^n)dx={\int }_{-1}^1(C_0+C_2{x}^2+C_4{x}^4+\cdots )dx+{\int }_{-1}^1(C_{1}x+C_3{x}^3+\cdots )dx=2{\int }_0^1(C_0+C_2{x}^2+C_4{x}^4+\cdots )dx+0$
(By Prop. of definite integral)
(since second integral contains odd function)
${\frac{(1+x{)}^{n+1}}{n+1}]}_{-1}^1={=2(C_{0}x+\frac{C_2{x}^3}{3}+\frac{C_4{x}^5}{5}+\cdots )]}_0^1\frac{2^{n+1}}{n+1}=2(C_0+\frac{C_2}{3}+\frac{C_4}{5}+\cdots )$
Hence $C_0+\frac{C_2}{3}+\frac{C_4}{5}+\cdots =\frac{2^n}{n+1}$