Sequence And Series Question 40
Question: $ \frac{(a-1)-\frac{{{(a-1)}^{2}}}{2}+\frac{{{(a-1)}^{3}}}{3}-….\infty }{(b-1)-\frac{{{(b-1)}^{2}}}{2}+\frac{{{(b-1)}^{3}}}{3}-…..\infty }= $
Options:
A) $ {\log_{b}}a $
B) $ {\log_{a}}b $
C) $ {\log_{e}}a-{\log_{e}}b $
D) $ {\log_{e}}a+{\log_{e}}b $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{(a-1)-\frac{{{(a-1)}^{2}}}{2}+\frac{{{(a-1)}^{3}}}{3}-…….\infty }{(b-1)-\frac{{{(b-1)}^{2}}}{2}+\frac{{{(b-1)}^{3}}}{3}-……\infty } $ $ =\frac{{\log_{e}}(1+\overline{a-1})}{{\log_{e}}(1+\overline{b-1})}=\frac{{\log_{e}}a}{{\log_{e}}b}={\log_{b}}a $ .
BETA
