_ r = 1 ^ ^ - 1 ( 3 r^2 - r + 9 ) is- - Vidyadip

MCQ

$\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

A
$\frac{\pi }{3}$
B
$\frac{\pi }{6}$
✓
$\frac{\pi }{2}$
D
$\frac{\pi }{12}$

✓

Answer

Correct option: C.

$\frac{\pi }{2}$

c
$T_{r}=\tan ^{-1}\left(\frac{\frac{1}{3}}{1+\frac{r}{3} \frac{(r-1)}{3}}\right)$

$\mathrm{T}_{\mathrm{r}}=\tan ^{-1}\left(\frac{\mathrm{r}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{r}-1)}{3}\right)$

$\mathrm{T}_{1}=\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)$

$\mathrm{T}_{2}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{1}{3}\right)$

$\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\mathrm{n}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{n}-1)}{3}\right)$

$S_{n}=\tan ^{-1}\left(\frac{n}{3}\right)=\frac{\pi}{2}$

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