Let S_{k} = sum_{r=1}^{k} tan^{-1}left(frac{6 | vedclass

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(C) We have $S_{k} = \sum_{r=1}^{k} 	an^{-1}\left(\frac{6^{r}}{2 \cdot 2^{2r} + 3 \cdot 3^{2r}}ight)$.Dividing the numerator and denominator by $3^

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

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(C) We have $S_{k} = \sum_{r=1}^{k} 	an^{-1}\left(\frac{6^{r}}{2 \cdot 2^{2r} + 3 \cdot 3^{2r}}ight)$.Dividing the numerator and denominator by $3^{2r}$,we get:$S_{k} = \sum_{r=1}^{k} 	an^{-1}\left(\frac{\frac{2^r}{3^r} \cdot 3^r}{2 \cdot \frac{2^{2r}}{3^{2r}} + 3}ight) = \sum_{r=1}^{k} 	an^{-1}\left(\frac{3^r \cdot (2/3)^r}{2 \cdot (2/3)^{2r} + 3}ight)$.Actually,rewrite the term inside $	an^{-1}$ as:$\frac{6^r}{2 \cdot 4^r + 3 \cdot 9^r} = \frac{6^r / 9^r}{2 \cdot (4/9)^r + 3} = \frac{(2/3)^r}{2 \cdot (2/3)^{2r} + 3}$.This can be written as $\frac{(2/3)^r / 3}{1 + \frac{2}{3} (2/3)^{2r}} = \frac{(2/3)^r - (2/3)^{r+1}}{1 + (2/3)^r \cdot (2/3)^{r+1}}$.Thus,the sum is a telescoping series:$S_{k} = \sum_{r=1}^{k} \left( 	an^{-1}((2/3)^r) - 	an^{-1}((2/3)^{r+1}) ight)$.$S_{k} = \left( 	an^{-1}(2/3) - 	an^{-1}((2/3)^2) ight) + \left( 	an^{-1}((2/3)^2) - 	an^{-1}((2/3)^3) ight) + \dots + \left( 	an^{-1}((2/3)^k) - 	an^{-1}((2/3)^{k+1}) ight)$.$S_{k} = 	an^{-1}(2/3) - 	an^{-1}((2/3)^{k+1})$.Taking the limit as $k ightarrow \infty$,$(2/3)^{k+1} ightarrow 0$.$S_{\infty} = 	an^{-1}(2/3) - 	an^{-1}(0) = 	an^{-1}(2/3) = \cot^{-1}(3/2)$.

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

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