Question

The value of $(\frac{{}^{50}{C}_0}{1}+\frac{{}^{50}{C}_2}{3}+\frac{{}^{50}{C}_4}{5}+....$ $+\frac{{}^{50}{C}_{50}}{51})$ is

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

Answer 1

Answer: (A)

$\left(\frac{{}^{50}{C}_0}{1}+\frac{{}^{50}{C}_2}{3}+\frac{{}^{50}{C}_4}{5}+....+\frac{{}^{50}{C}_{50}}{51}\right)$
$=\frac{1}{1}+\frac{50\times 49}{3\times 2!}+\frac{50\times 49\times 48\times 47}{5\times 4!}+....$
$=\frac{1}{51}\left(51+\frac{51\times 50\times 49}{3!}+\frac{\begin{array}{rl}& 51\times 50\times 49\\ & \times 48\times 47\end{array}}{5!}+....\right)$
$=\frac{1}{51}{(}^{51}{C}_1{+}^{51}{C}_3{+}^{51}{C}_5+....)$
$=\frac{1}{51}{.2}^{51-1}=\frac{2^{50}}{51}$