Sequence And Series Question 573
Question: $ (1+3){\log_{e}}3+\frac{1+3^{2}}{2,!}{{({\log_{e}}3)}^{2}}+\frac{1+3^{3}}{3,!}{{({\log_{e}}3)}^{3}}+…..\infty = $
[Roorkee 1989]
Options:
A) 28
B) 30
C) 25
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ S={\log_{e}}3+\frac{{{({\log_{e}}3)}^{2}}}{2!}+\frac{{{({\log_{e}}3)}^{3}}}{3!}+……..+3{\log_{e}}3+\frac{{{(3{\log_{e}}3)}^{2}}}{2!}+\frac{{{(3{\log_{e}}3)}^{3}}}{3!}+…. $ $ =({e^{{\log_{e}}3}}-1)+({e^{3{\log_{e}}3}}-1)=(3-1)+(3^{3}-1)=28 $ .
BETA
