MCQ
$\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ મેળવો.
- A$\frac{\pi }{3}$
- B$\frac{\pi }{6}$
- ✓$\frac{\pi }{2}$
- D$\frac{\pi }{12}$
$\mathrm{T}_{\mathrm{r}}=\tan ^{-1}\left(\frac{\mathrm{r}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{r}-1)}{3}\right)$
$\mathrm{T}_{1}=\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)$
$\mathrm{T}_{2}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{1}{3}\right)$
$\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\mathrm{n}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{n}-1)}{3}\right)$
$S_{n}=\tan ^{-1}\left(\frac{n}{3}\right)=\frac{\pi}{2}$
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