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Question

Mathematics Question on Binomial theorem

C01+C23+C45+C67+........=\frac{C_{0}}{1}+\frac{C_{2}}{3}+\frac{C_{4}}{5}+\frac{C_{6}}{7}+ ........ =

A

2n+1n+1\frac{2^{n+1}}{n+1}

B

2n+1−1n+1\frac{2^{n+1}-1}{n+1}

C

2nn+1\frac{2^{n}}{n+1}

D

None of these

Answer

2nn+1\frac{2^{n}}{n+1}

Explanation

Solution

Putting the value of C0,C2,C4.....C_0, C_2, C_4....., we get =1+n(n+1)3.2!+n(n−1)(n−2)(n−3)5.4!+.....=1n+1= 1 +\frac{n\left(n+1\right)}{3.2!} +\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5.4!} +..... = \frac{1}{n+1} [(n+1)+(n+1)n(n−1)3!+(n+1)n(n−1)(n−2)(n−3)5!+.....]\left[\left(n+1\right)+ \frac{\left(n+1\right)n\left(n-1\right)}{3!} + \frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5!}+.....\right] Put n+1=Nn + 1 = N =1N[N+N(N−1)(N−2)3!+N(N−1)(N−2)(N−3)(N−4)5!+.....]= \frac{1}{N} \left[ N +\frac{N\left(N-1\right)\left(N-2\right)}{3!}+\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)\left(N-4\right)}{5!}+.....\right] = \frac{1}{N} \left\\{^{N}C_{1}+^{N}C_{3}+^{N}C_{5} + .....\right\\} = \frac{1}{N} \left\\{2^{N-1}\right\\} = \frac{2^{n}}{n+1}\quad\left\\{\because N = n+1\right\\}