Inverse Trigonometric Functions Question 68
Question: $ \sum\limits_{m=1}^{n}{{{\tan }^{-1}}}( \frac{2m}{m^{4}+m^{2}+2} ) $ is equal to
Options:
A) $ {{\tan }^{-1}}( \frac{n^{2}+n}{n^{2}+n+2} ) $
B) $ {{\tan }^{-1}}( \frac{n^{2}-n}{n^{2}-n+2} ) $
C) $ {{\tan }^{-1}}( \frac{n^{2}+n+2}{n^{2}+n} ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \sum\limits_{m=1}^{n}{{{\tan }^{-1}}( \frac{2m}{m^{4}+m^{2}+2} )} $
$ =\sum\limits_{m=1}^{n}{{{\tan }^{-1}}( \frac{2m}{1+(m^{2}+m+1)(m^{2}-m+1)} )} $
$ =\sum\limits_{m=1}^{n}{{{\tan }^{-1}}( \frac{(m^{2}+m+1)-(m^{2}-m+1)}{1+(m^{2}+m+1)(m^{2}-m+1)} )} $ = $ \sum\limits_{m=1}^{n}{[{{\tan }^{-1}}(m^{2}+m+1)-{{\tan }^{-1}}(m^{2}-m+1)]} $
$ =({{\tan }^{-1}}3-{{\tan }^{-1}}1)+({{\tan }^{-1}}7-{{\tan }^{-1}}3)+ $
$ ({{\tan }^{-1}}13-{{\tan }^{-1}}7)+……+[{{\tan }^{-1}}(n^{2}+n+1) $
$ -{{\tan }^{-1}}(n^{2}-n+1)] $ = $ {{\tan }^{-1}}(n^{2}+n+1)-{{\tan }^{-1}}1 $ = $ {{\tan }^{-1}}( \frac{n^{2}+n}{2+n^{2}+n} ) $ .
BETA
