Binomial Theorem And Its Simple Applications Question 89
Question: $ \frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+…. $ =
[RPET 1999]
Options:
A) $ \frac{{2^{n+1}}}{n+1} $
B) $ \frac{{2^{n+1}}-1}{n+1} $
C) $ \frac{2^{n}}{n+1} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Putting the values of $ C_0,C_2,C_4…., $ we get $ =1+\frac{n(n-1)}{3.2!}+\frac{n(n-1)(n-2)(n-3)}{5.4!}+…. $ = $ \frac{1}{n+1}[ (n+1)+\frac{(n+1)n(n-1)}{3!}+\frac{(n+1)n(n-1)(n-2)(n-3)}{5!}+…. ] $
Put $ n+1 $ =N = $ \frac{1}{N}[ N+\frac{N(N-1)(N-2)}{3!}+\frac{N(N-1),(N-2)(N-3)(N-4)}{5!}+…. ] $
$ =\frac{1}{N}{ {{,}^{N}}C_1+{{,}^{N}}C_3+{{,}^{N}}C_5+…. } $
$ =\frac{1}{N}{ {2^{N-1}} }=\frac{2^{n}}{n+1} $
$ {\because N=n+1} $
Trick: Put n=1, then $ S_1=\frac{^{1}C_0}{1}=\frac{1}{1}=1 $ At n=2, $ S_2=\frac{^{2}C_0}{1}+\frac{^{2}C_2}{3}=1+\frac{1}{3}=\frac{4}{3} $ Also (c)
$ \Rightarrow ,S_1=1,S_2=\frac{4}{3} $
BETA
