- r -1 n tan -12r-1-1-22 -1 is equal toA- tan -12 n B- tan -12 n -4C- tan -12 n -1D- tan -12 n -1-4

The correct option is B tan^ 1 (2^n) π/4∑ tan^ 1 ( 2^r 2^r 1/1+2^r. 2^r 1 ) =∑^n_r=1 [tan^ 1 (2^r) tan^ 1(2^r 1)] =tan^ 1(2^n) tan^ 1(1) =tan^ 1(2^n) ...

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Byju's Answer

Standard XII

Mathematics

Functions

∑ r =1 n tan ...

Question

$\sum_{r = 1}^{n} t a n^{- 1} (\frac{2^{r - 1}}{1 + 2^{2 r - 1}})$ is equal to

t a n^{- 1} (2^{n})

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t a n^{- 1} (2^{n}) - \frac{π}{4}

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t a n^{- 1} (2^{n + 1})

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t a n^{- 1} (2^{n + 1}) - \frac{π}{4}

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Solution

The correct option is B $t a n^{- 1} (2^{n}) - \frac{π}{4}$
$\sum t a n^{- 1} (\frac{2^{r} - 2^{r - 1}}{1 + 2^{r} . 2^{r - 1}})$
$= \sum_{r = 1}^{n} [t a n^{- 1} (2^{r}) - t a n^{- 1} (2^{r - 1})]$
$= t a n^{- 1} (2^{n}) - t a n^{- 1} (1)$
$= t a n^{- 1} (2^{n}) - \frac{π}{4}$

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Q. Find the sum of the following series:

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Q. Solve the following:
(a)

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(b)

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In a $Δ A B C$ of A $- t a n^{- 1}$ and B $= t a n^{- 1} 3$ then C is equal to

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∑nr=1tan−1(2r−11+22r−1) is equal to

The correct option is B tan−1(2n)−π4∑tan−1(2r−2r−11+2r.2r−1) =∑nr=1[tan−1(2r)−tan−1(2r−1)] =tan−1(2n)−tan−1(1) =tan−1(2n)−π4

$\sum_{r = 1}^{n} t a n^{- 1} (\frac{2^{r - 1}}{1 + 2^{2 r - 1}})$ is equal to

The correct option is B $t a n^{- 1} (2^{n}) - \frac{π}{4}$
$\sum t a n^{- 1} (\frac{2^{r} - 2^{r - 1}}{1 + 2^{r} . 2^{r - 1}})$
$= \sum_{r = 1}^{n} [t a n^{- 1} (2^{r}) - t a n^{- 1} (2^{r - 1})]$
$= t a n^{- 1} (2^{n}) - t a n^{- 1} (1)$
$= t a n^{- 1} (2^{n}) - \frac{π}{4}$