Binomial Theorem And Its Simple Applications Question 103
Question: $ \frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+….+\frac{C_{n}}{n+1}= $
[RPET 1996]
Options:
A) $ \frac{2^{n}}{n+1} $
B) $ \frac{2^{n}-1}{n+1} $
C) $ \frac{{2^{n+1}}-1}{n+1} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Proceeding as above and putting n+1=N. So given term can be written as $ \frac{1}{N}{ {{,}^{N}}C_1+{{,}^{N}}C_2+{{,}^{N}}C_3+…. } $ = $ \frac{1}{N}{ 2^{N}-1 }=\frac{1}{n+1}({2^{n+1}}-1) $
$ (\because N=n+1) $
BETA
