If a=2, then sum of series
$c o t^{- 1} (2 a^{- 1} + a) + c o t^{- 1} (2 a^{- 1} + 3 a) + c o t^{- 1} (2 a^{- 1} + 6 a) + c o t^{- 1} (2 a^{- 1} + 10 a) +$ .....upto infinite terms, is.

\frac{π}{4}

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{6}

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Solution

The correct option is A $\frac{π}{4}$
$a = 2$
General term $= {cot}^{- 1} ({2 a}^{- 1} + \frac{r (r + 1) a}{2})$
$= {cot}^{- 1} (1 + r (r + 1))$
$= {tan}^{- 1} (\frac{1}{(1 + r (r + 1))})$
$= {tan}^{- 1} (\frac{(r + 1) - r}{(1 + r (r + 1))})$
$= {tan}^{- 1} (r + 1) - {tan}^{- 1} (r)$
$∴ \sum_{r = 1}^{\infty} {cot}^{- 1} (2 a^{- 1} + \frac{r (r + 1) a}{2})$
$= \sum_{r = 1}^{\infty} {tan}^{- 1} (r + 1) - {tan}^{- 1} (r)$
$= {tan}^{- 1} (2) - {tan}^{- 1} (1) + {tan}^{- 1} (3) - {tan}^{- 1} (2) . . . . . {tan}^{- 1} (\infty)$
$\Rightarrow - \frac{π}{4} + \frac{π}{2} = \frac{π}{4} .$
Hence the answer is $\frac{π}{4} .$

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Similar questions

c o t^{- 1} (1^{2} + \frac{3}{4}) + c o t^{- 1} (2^{2} + \frac{3}{4}) + c o t^{- 1} (3^{2} + \frac{3}{4}) + . . . . . . .

c o t^{- 1} 3 + c o t^{- 1} 7 + c o t^{- 1} 13 + . . . .

{cot}^{- 1} (1^{2} + \frac{3}{4}) + {cot}^{- 1} (2^{2} + \frac{3}{4}) + {cot}^{- 1} (3^{2} + \frac{3}{4}) + \dots

c o t^{- 1} (1^{2} + \frac{3}{4}) + c o t^{- 1} (2^{2} + \frac{3}{4}) + c o t^{- 1} (3^{2} + \frac{3}{4}) + . . . . . . .

a > 3

A, B, C

△ A B C,

\sqrt{2 a^{2} - 9} cot A + 2 a cot B + \sqrt{2 a^{2} + 9} cot C = 12 a,

{cot}^{2} A + {cot}^{2} B + {cot}^{2} C =

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