Sequence And Series Question 50
Question: If $ b=a-\frac{a^{2}}{2}+\frac{a^{3}}{3}-\frac{a^{4}}{4}+.. $ then $ b+\frac{b^{2}}{2,!}+\frac{b^{3}}{3,!}+\frac{b^{4}}{4,!}+…\infty = $
Options:
A) $ {\log_{e}}a $
B) $ {\log_{e}}b $
C) $ a $
D) $ e^{a} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ b={\log_{e}}(1+a)\Rightarrow 1+a=e^{b} $
$ \Rightarrow ,1+a=1+\frac{b}{1!}+\frac{b^{2}}{2!}+…. $
Þ $ a=b+\frac{b^{2}}{2,!}+….. $
BETA
