Sequence And Series Question 80
Question: In $ n=(1999),! $ then $ \sum\limits_{x=1}^{1999}{{\log_{n}}x} $ is equal to
[AMU 2002]
Options:
A) 1
B) 0
C) $ \sqrt[1999]{1999} $
D) - 1
Show Answer
Answer:
Correct Answer: A
Solution:
$ \sum\limits_{x=1}^{1999}{{\log_{n}}x} $ $ ={\log_{(1999)!}}1+{\log_{(1999)!}}2+…….+{\log_{(1999)!}}1999 $ $ ={\log_{(1999)!}}(1.2.3…….1999)={\log_{(1999)!}}(1999)!=1 $ .
BETA
