Question

The sum of the binomial coefficients $\left(\frac{{ }^{50} C_{0}}{1}+\frac{{ }^{50} C_{2}}{3}+\frac{{ }^{50} C_{4}}{5}+\ldots+\frac{{ }^{50} C_{50}}{51}\right)$ is equal to

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

Answer 1

Answer: (A)

$\left(\frac{{ }^{50} C_{0}}{1}+\frac{{ }^{50} C_{2}}{3}+\frac{{ }^{50} C_{4}}{5}+\ldots+\frac{{50} C_{50}}{51}\right)$
$=\frac{1}{51}\left({ }^{51} C_{1}+{ }^{51} C_{3}+{ }^{51} C_{5}+\ldots\right)$
$=\frac{1}{51} \cdot 2^{51-1}=\frac{2^{50}}{51}$