Binomial Theorem And Its Simple Applications Question 118
Question: $ \frac{1}{1!(n-1),!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+….= $
[AMU 2005]
Options:
A) $ \frac{2^{n}}{n!} $ ; for all even values of n
B) $ \frac{{2^{n-1}}}{n!} $ ; for all values of n i.e., all even odd values
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Multiplying each term by n! the question reduces to $ \frac{n!}{1!(n-1)!}+\frac{1}{3!}.\frac{n!}{(n-3),!}+\frac{1}{5!}.\frac{n!}{(n-5)!}+…. $
$ ={{,}^{n}}C_1+{{,}^{n}}C_3+{{,}^{n}}C_5+….={2^{n-1}} $ . Thus $ \frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+…. $
$ =\frac{1}{n!}{2^{n-1}} $ .
BETA
