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Mathematics Question on Combinations

The value of (50C01+50C23+50C45+....\left( \frac{^{50}{{C}_{0}}}{1}+\frac{^{50}{{C}_{2}}}{3}+\frac{^{50}{{C}_{4}}}{5}+.... \right. +50C5051)\left. +\frac{^{50}{{C}_{50}}}{51} \right) is

A

25051\frac{{{2}^{50}}}{51}

B

250−151\frac{{{2}^{50}}-1}{51}

C

250−150\frac{{{2}^{50}}-1}{50}

D

251−151\frac{{{2}^{51}}-1}{51}

Answer

25051\frac{{{2}^{50}}}{51}

Explanation

Solution

(50C01+50C23+50C45+....+50C5051)\left( \frac{^{50}{{C}_{0}}}{1}+\frac{^{50}{{C}_{2}}}{3}+\frac{^{50}{{C}_{4}}}{5}+....+\frac{^{50}{{C}_{50}}}{51} \right)
=11+50×493×2!+50×49×48×475×4!+....=\frac{1}{1}+\frac{50\times 49}{3\times 2!}+\frac{50\times 49\times 48\times 47}{5\times 4!}+....
=151(51+51×50×493!+51×50×49 ×48×47 5!+....)=\frac{1}{51}\left( 51+\frac{51\times 50\times 49}{3!}+\frac{\begin{aligned} & 51\times 50\times 49 \\\ & \times 48\times 47 \\\ \end{aligned}}{5!}+.... \right)
=151(51C1+51C3+51C5+....)=\frac{1}{51}{{(}^{51}}{{C}_{1}}{{+}^{51}}{{C}_{3}}{{+}^{51}}{{C}_{5}}+....)
=151.251−1=25051=\frac{1}{51}{{.2}^{51-1}}=\frac{{{2}^{50}}}{51}