Sequence And Series Question 175
Question: $ 1+\frac{{{({\log_{e}}n)}^{2}}}{2,!}+\frac{{{({\log_{e}}n)}^{4}}}{4,!}+….= $
[MP PET 1996]
Options:
A) $ n $
B) $ 1/n $
C) $ \frac{1}{2}(n+{n^{-1}}) $
D) $ \frac{1}{2}(e^{n}+{e^{-n}}) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1+\frac{{{({\log_{e}}n)}^{2}}}{2,!}+\frac{{{({\log_{e}}n)}^{4}}}{4,!}+….=\frac{{e^{{\log_{e}}n}}+{e^{-{\log_{e}}n}}}{2} $ = $ \frac{n+{n^{-1}}}{2} $ .
BETA
