r=1∑ntan−1(1+22r−12r−1) is equal to :

r=1∑ntan−1(1+22r−12r−1)=r=1∑ntan−1(1+2r.2r−12r−1(2−1)) =r=1∑ntan−1(1+2r.2r−12r−2r−1) =r=1∑ntan−12r−tan−1(2r−1) =tan−12−tan−11+tan−122−tan−121+⋯+tan−12n−tan−12n−1 =tan−1(2n)−4π

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_{ r=1 }^{ n }{ { tan }^{ -1 } } ( { { 2 }^{ r-1 } }{ 1+{ 2

Question

$r = 1 \sum n t an^{- 1} (\frac{2 ^{r - 1}}{1 + 2 ^{2 r - 1}})$ is equal to :

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

t an^{- 1} (2^{n}) - \frac{π}{4}

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

t an^{- 1} (2^{n + 1}) - \frac{π}{4}

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Chloe Discussed

r = 1 \sum n t an^{- 1} (\frac{2 ^{r - 1}}{1 + 2 ^{2 r - 1}})

is equal to :

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6 mins ago

Text solutionVerified

r = 1 \sum n tan^{- 1} (\frac{2 ^{r - 1}}{1 + 2 ^{2 r - 1}}) = r = 1 \sum n tan^{- 1} (\frac{2 ^{r - 1} ( 2 - 1 )}{1 + 2 ^{r} . 2 ^{r - 1}})

= r = 1 \sum n tan^{- 1} (\frac{2 ^{r} - 2 ^{r - 1}}{1 + 2 ^{r} . 2 ^{r - 1}})

= r = 1 \sum n tan^{- 1} 2^{r} - tan^{- 1} (2^{r - 1})

= tan^{- 1} 2 - tan^{- 1} 1 + tan^{- 1} 2^{2} - tan^{- 1} 2^{1} + \dots + tan^{- 1} 2^{n} - tan^{- 1} 2^{n - 1}

= tan^{- 1} (2^{n}) - \frac{π}{4}

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Question Text	$r = 1 \sum n t an^{- 1} (\frac{2 ^{r - 1}}{1 + 2 ^{2 r - 1}})$ is equal to :
Answer Type	Text solution:1
Upvotes	150

r=1∑n​tan−1(1+22r−12r−1​) is equal to :

Text solutionVerified

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$r = 1 \sum n t an^{- 1} (\frac{2 ^{r - 1}}{1 + 2 ^{2 r - 1}})$ is equal to :