Let S k = r = 1 k tan - 1 ( 6 r 2 2 r + 1 + 3 2 r + 1 ) . Then lim k S k is equal to

Question

$Let S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{6^{r}}{2^{2 r + 1} + 3^{2 r + 1}}) . Then \lim_{k \to \infty} S_{k} is equal to$

Smart Achievers · Accepted Answer

(3)

$S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{\frac{1}{3} {(\frac{2}{3})}^{r}}{{(\frac{2}{3})}^{2 r + 1} + 1}) = \sum_{r = 1}^{k} [\tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} {(\frac{2}{3})}^{r + 1}] = \tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} (0) = \cot^{- 1} (\frac{3}{2})$

Solution

Q.
$Let S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{6^{r}}{2^{2 r + 1} + 3^{2 r + 1}}) . Then \lim_{k \to \infty} S_{k} is equal to$

1 $\tan^{- 1} (\frac{3}{2})$

2 $\frac{π}{2}$

3 $\cot^{- 1} (\frac{3}{2})$

4 $\tan^{- 1} (3)$

Ans.
(3)

$S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{\frac{1}{3} {(\frac{2}{3})}^{r}}{{(\frac{2}{3})}^{2 r + 1} + 1}) = \sum_{r = 1}^{k} [\tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} {(\frac{2}{3})}^{r + 1}] = \tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} (0) = \cot^{- 1} (\frac{3}{2})$

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Solution

Q. Let Sk=∑r=1ktan-1(6r22r+1+32r+1). Then limk→∞Sk is equal to

1 tan-1(32)

2 π2

3 cot-1(32)

4 tan-1(3)

Ans. (3) Sk=∑r=1ktan-1(13(23)r(23)2r+1+1)=∑r=1k[tan-1(23)r-tan-1(23)r+1]=tan-1(23)r-tan-1(0)=cot-1(32)