Sequence And Series Question 571
Question: $ 1+\frac{{\log_{e}}x}{1,!}+\frac{{{({\log_{e}}x)}^{2}}}{2,!}+\frac{{{({\log_{e}}x)}^{3}}}{3,!}+…..\infty = $
[Kurukshetra CEE 1998; JMI CET 2000]
Options:
A) $ {\log_{e}}x $
B) $ x $
C) $ {x^{-1}} $
D) $ -{\log_{e}}(1+x) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 1+\frac{{\log_{e}}x}{1,!}+\frac{{{({\log_{e}}x)}^{2}}}{2\ !}+\frac{{{({\log_{e}}x)}^{3}}}{3\ !}+……+\frac{{{({\log_{e}}x)}^{n}}}{n\ !}+…. $ $ ={e^{({\log_{e}}x)}}=x $ .
BETA
