If S - tan - 1 1 - n 2 - n - 1 - tan - 1 1 - n 2 - 3 n - 3 - - - tan - 1 1 - 1 - n - 19 n - 20 then tan S is equal to

The correct option is C 20 / n ^ 2 + 20 n + 1 [ S=tan ^ 1 # ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If S = tan ...

Question

If $S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + \dots + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)})$ then $tan S$ is equal to

\frac{20}{401 + 20 n}

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\frac{n}{n^{2} + 20 n + 1}

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\frac{20}{n^{2} + 20 n + 1}

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\frac{n}{401 + 20 n}

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Solution

The correct option is C $\frac{20}{n^{2} + 20 n + 1}$
$\begin{matrix} S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + . . . . + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)}) S = {tan}^{- 1} (\frac{n + 1 - n}{1 + n (n + 1)}) + {tan}^{- 1} [\frac{(n + 2) - (n - 1)}{1 + (n + 1) (n + 2)}] + . . . . . + {tan}^{- 1} [\frac{(n + 20) - (n + 19)}{1 + (n + 19) (n + 20)}] = [{tan}^{- 1} (n + 1) - {tan}^{- 1} (n)] + [{tan}^{- 1} (n + 2) - {tan}^{- 1} (n + 1)] + . . . . . . + [{tan}^{- 1} (n + 20) - {tan}^{- 1} (n + 19)] = {tan}^{- 1} (n + 20) - t a n^{- 1} (n) = {tan}^{- 1} (\frac{20}{1 + n^{2} + 20 n}) t a n s = \frac{20}{n^{2} + 20 n + 1} H e n c e, t h e o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + . . . . . + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)})

, then

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If S=tan−1(1n2+n+1)+tan−1(1n2+3n+3)+…+tan−1(11+(n+19)(n+20)) then tanS is equal to

If $S = {tan}^{- 1} (\frac{1}{n^{2} + n + 1}) + {tan}^{- 1} (\frac{1}{n^{2} + 3 n + 3}) + \dots + {tan}^{- 1} (\frac{1}{1 + (n + 19) (n + 20)})$ then $tan S$ is equal to