Binomial Theorem And Its Simple Applications Question 22
Question: The value of $ \frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+….. $ is equal to
Options:
A) $ \frac{2^{n}+1}{n+1} $
B) $ \frac{2^{n}}{n+1} $
C) $ \frac{2^{n}+1}{n-1} $
D) $ \frac{2^{n}-1}{n+1} $
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Answer:
Correct Answer: D
Solution:
- [d] $ \frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+….=\frac{n}{2}+\frac{n(n-1)(n-2)}{4!} $
$ +\frac{n(n-1)(n-2)(n-3)(n-4)}{6!}+…. $
$ =\frac{1}{n+1}[ \frac{(n+1)n}{2!}+\frac{(n+1)n(n-1)(n-2)}{4!}+ . $
$ . \frac{(n+1)n(n-1)(n-2)(n-3)(n-4)}{6!}+…. ] $
$ =\frac{1}{n+1}{{[}^{n+1}}C_2+{{,}^{n+1}}C_4+{{,}^{n+1}}C_6+….] $
$ =\frac{1}{n+1}[ {2^{n+1-1}}{{-}^{n+1}}C_0 ]=\frac{2^{n}-1}{n+1} $
BETA
