Question:
$\frac{C_{0}}{1}+\frac{C_{2}}{3}+\frac{C_{4}}{5}+\frac{C_{6}}{7}+ ........ =$
$\frac{C_{0}}{1}+\frac{C_{2}}{3}+\frac{C_{4}}{5}+\frac{C_{6}}{7}+ ........ =$
Updated On: Jul 6, 2022
- $\frac{2^{n+1}}{n+1}$
- $\frac{2^{n+1}-1}{n+1}$
- $\frac{2^{n}}{n+1}$
- None of these
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The Correct Option is C
Solution and Explanation
Putting the value of $C_0, C_2, C_4.....$, we get
$= 1 +\frac{n\left(n+1\right)}{3.2!} +\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5.4!} +..... = \frac{1}{n+1}$
$\left[\left(n+1\right)+ \frac{\left(n+1\right)n\left(n-1\right)}{3!} + \frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{5!}+.....\right]$
Put $n + 1 = N$
$= \frac{1}{N} \left[ N +\frac{N\left(N-1\right)\left(N-2\right)}{3!}+\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)\left(N-4\right)}{5!}+.....\right]$
$= \frac{1}{N} \left\{^{N}C_{1}+^{N}C_{3}+^{N}C_{5} + .....\right\}$
$= \frac{1}{N} \left\{2^{N-1}\right\} = \frac{2^{n}}{n+1}\quad\left\{\because N = n+1\right\}$
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.