Question

The value of $ \left( \frac{^{50}{{C}_{0}}}{1}+\frac{^{50}{{C}_{2}}}{3}+\frac{^{50}{{C}_{4}}}{5}+.... \right. $ $ \left. +\frac{^{50}{{C}_{50}}}{51} \right) $ is

• A. $ \frac{{{2}^{50}}}{51} $
• B. $ \frac{{{2}^{50}}-1}{51} $
• C. $ \frac{{{2}^{50}}-1}{50} $
• D. $ \frac{{{2}^{51}}-1}{51} $

Answer 1

The Correct Option is A

$ \left( \frac{^{50}{{C}_{0}}}{1}+\frac{^{50}{{C}_{2}}}{3}+\frac{^{50}{{C}_{4}}}{5}+....+\frac{^{50}{{C}_{50}}}{51} \right) $
$=\frac{1}{1}+\frac{50\times 49}{3\times 2!}+\frac{50\times 49\times 48\times 47}{5\times 4!}+.... $
$=\frac{1}{51}\left( 51+\frac{51\times 50\times 49}{3!}+\frac{\begin{align} & 51\times 50\times 49 \\ & \times 48\times 47 \\ \end{align}}{5!}+.... \right) $
$=\frac{1}{51}{{(}^{51}}{{C}_{1}}{{+}^{51}}{{C}_{3}}{{+}^{51}}{{C}_{5}}+....) $
$=\frac{1}{51}{{.2}^{51-1}}=\frac{{{2}^{50}}}{51} $