Sequence And Series Question 562
Question: $ \frac{2}{1,!}{\log_{e}}2+\frac{2^{2}}{2,!}{{({\log_{e}}2)}^{2}}+\frac{2^{3}}{3,!}{{({\log_{e}}2)}^{3}}+…..\infty = $
Options:
A) 2
B) 3
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Sum = $ ({e^{2{\log_{e}}2}}-1)=4-1=3 $ .
BETA
