Question:
\( \sum_{m=1}^{n} \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \) is equal to
\( \sum_{m=1}^{n} \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \) is equal to
Show Hint
If expression looks complex, try converting into telescoping form.
Updated On: Apr 23, 2026
- $\tan^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
- $\tan^{-1}\left(\frac{n^2-n}{n^2-n+2}\right)$
- $\tan^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$
- None of these
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Use identity:
\[
\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)
\]
Step 1: Rewrite general term.
\[ \tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right) = \tan^{-1}(m^2+m+1) - \tan^{-1}(m^2-m+1) \]
Step 2: Apply telescoping sum.
Terms cancel successively.
Step 3: Evaluate limits.
\[ = \tan^{-1}(n^2+n+1) - \tan^{-1}(1) \]
Step 4: Simplify expression.
Final simplifies to: \[ \tan^{-1}\left(\frac{n^2+n}{n^2+n+2}\right) \] Conclusion:
Answer = Option (A)
Step 1: Rewrite general term.
\[ \tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right) = \tan^{-1}(m^2+m+1) - \tan^{-1}(m^2-m+1) \]
Step 2: Apply telescoping sum.
Terms cancel successively.
Step 3: Evaluate limits.
\[ = \tan^{-1}(n^2+n+1) - \tan^{-1}(1) \]
Step 4: Simplify expression.
Final simplifies to: \[ \tan^{-1}\left(\frac{n^2+n}{n^2+n+2}\right) \] Conclusion:
Answer = Option (A)
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