Sequence And Series Question 327
Question: $ 1+x{\log_{e}}a+\frac{x^{2}}{2,!}{{({\log_{e}}a)}^{2}}+\frac{x^{3}}{3,!}{{({\log_{e}}a)}^{3}}+…= $
[EAMCET 2002]
Options:
A) $ a^{x} $
B) x
C) $ {a^{{\log_{a}}x}} $
D) a
Show Answer
Answer:
Correct Answer: A
Solution:
Series = $ 1+x{\log_{e}}a+\frac{x^{2}}{2!}{{[ {\log_{e}}a ]}^{2}}+\frac{x^{3}}{3!}{{[{\log_{e}}a]}^{3}}+… $ $ ={e^{x{\log_{e}}a}}={e^{{\log_{e}}a^{x}}}=a^{x} $ .
BETA
