Question

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

• A. $\frac{2^{50}}{51}$
• B. $\frac{2^{50}-1}{51}$
• C. $\frac{2^{50}-1}{50}$
• D. None of these

Answer 1

The correct option is A $\frac{2^{50}}{51}$

$(\frac{{}^{50}{C}_0}{1}+\frac{{}^{50}{C}_2}{3}+...+\frac{{}^{50}{C}_{50}}{51})$
$=(\frac{1}{1}+\frac{50\times 49}{3\times 2!}+...+\frac{1}{51})$
$=\frac{1}{51}(\frac{51}{1!}+\frac{51\times 50\times 49}{3!}+...+1)$
$=\frac{1}{51}({}^{51}{C}_1{+}^{51}{C}_3+...{+}^{51}{C}_{51})$
$=\frac{1}{51}{2}^{51-1}=\frac{2^{50}}{51}$