Binomial Theorem And Its Simple Applications Question 48
Question: $ \frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+….= $
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:
- [c] Putting the value of $ C_0,C_2C_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 } $
BETA
