MCQ
$\sum_{m=1}^n\ \ \tan^{-1} \left(\frac{2m}{m^4+m^2+2}\right)=.............. $
- ✓$\tan^{-1} \left(\frac{n^2+n}{n^2+n+2}\right)$
- B$\tan^{-1} \left(\frac{n^2+n}{n^2-n+2}\right)$
- C$\tan^{-1} \left(\frac{n^2+n+2}{n^2-n}\right)$
- Dઆમાંથી એક પણ નહિં.
$\sum_{m=1}^n\ \ \tan^{-1} \left(\frac{2m}{m^4+m^2+2}\right)$
$= \sum_{m=1}^{n} \tan^{-1} \left( \frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}\right)$
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