Sequence And Series Question 565
Question: If $ y=1+\frac{x}{1,!}+\frac{x^{2}}{2,!}+\frac{x^{3}}{3,!}+……\infty $ , then $ x= $
Options:
A) $ {\log_{e}}y $
B) $ {\log_{e}}\frac{1}{y} $
C) $ e^{y} $
D) $ {e^{-y}} $ !
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=1+\frac{x}{1\ !}+\frac{x^{2}}{2\ !}+\frac{x^{3}}{3\ !}+……=e^{x}\Rightarrow x={\log_{e}}y $ .
BETA
