Question

The value of $\frac{C_1}{2}$ + $\frac{C_3}{4}$ + $\frac{C_5}{6}$ + ...... is equal to

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

Answer 1

The correct option is A

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

We know that

$\frac{(1+x{)}^n-(1-x{)}^n}{2}$ = $C_{1}x+C_3{x}^3+C_5{x}^5+........$

Integrating from x = 0 to x = 1, we get

$\frac{1}{2}$${\int }_0^1(1+x{)}^n-(1-x{)}^{n}dx$

= ${\int }_0^1(C_{1}x+C_3{x}^3+C_5{x}^5+.......)dx$

⇒ $\frac{1}{2}{\{\frac{(1+x{)}^{n+1}}{n+1}+\frac{(1-x{)}^{n+1}}{n+1}\}}_0^1=\frac{C_1}{2}$ + $\frac{C_3}{4}$ + $\frac{C_5}{6}$ + ....

or $\frac{C_1}{2}$ + $\frac{C_3}{4}$ + $\frac{C_5}{6}$ + .......= $\frac{1}{2}$ {$\frac{2^{n+1}-1}{n+1}$ + $\frac{0-1}{n+1}$}

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