Binomial Theorem And Its Simple Applications Question 193
Question: The value of $ \sum\limits_{n=1}^{\infty }{\frac{^{n}C_0+…{{+}^{n}}C_{n}}{^{n}P_{n}}} $ is
[Kerala (Engg.) 2005]
Options:
A) $ e^{2} $
B) e
C) $ e^{2}-1 $
D) $ e-1 $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \sum\limits_{n=1}^{\infty }{\frac{^{n}C_0+…….+{{,}^{n}}C_{n}}{{{,}^{n}}P_{n}}} $
$ =\frac{{{,}^{1}}C_0+{{,}^{1}}C_1}{{{,}^{1}}P_1}+\frac{{{,}^{2}}C_0+{{,}^{2}}C_1+{{,}^{2}}C_2}{{{,}^{2}}P_2}+\frac{^{3}C_0+{{,}^{3}}C_1+{{,}^{3}}C_2+{{,}^{3}}C_3}{{{,}^{3}}P_3} $ +… $ =\frac{2^{1}}{1!}+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+……. $
$ ( 1+\frac{2}{1!}+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+……. )-1 $
$ =e^{2}-1 $ .
BETA
