Sequence And Series Question 81
Question: $ {e^{( x,-,\frac{1}{2}{{(x,-,1)}^{2}},+,\frac{1}{3}{{(x,-,1)}^{3}},-,\frac{1}{4}{{(x,-,1)}^{4}}+……. )}} $ is equal to
[DCE 2001]
Options:
A) $ \log x $
B) $ \log (x-1) $
C) x
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ {e^{( x,-,\frac{1}{2}{{(x,-,1)}^{2}},+,\frac{1}{3}{{(x,-,1)}^{3}}-……… )}} $ $ ={e^{( (x-1)-\frac{1}{2}{{(x-1)}^{2}}+\frac{1}{3}{{(x-1)}^{3}}-…… )+,1}} $ = $ {e^{\log (1+x-1)}}e={e^{\log x}}.e=xe. $
BETA
