Sequence And Series Question 544
Question: If $ S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n),!},} $ then $ S $ =
Options:
A) $ x+{x^{-1}} $
B) $ x-{x^{-1}} $
C) $ \frac{1}{2}(x+{x^{-1}}) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)!}=( \frac{{e^{\log x}}+{e^{-\log x}}}{2} )}=\frac{x+{x^{-1}}}{2} $ .
BETA
