Let $S_{k}=\sum_{r=1}^{k} \tan ^{-1}\left(\frac{6^{r}}{2^{2 r+1}+3^{2 r+1}}\right) .$ Then $\lim _{k \rightarrow \infty} S_{k}$ is equal to :
Correct Option: , 3
$\mathrm{S}_{\mathrm{k}}=\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{6^{\mathrm{r}}}{2^{2 \mathrm{r}+1}+3^{2 \mathrm{r}+1}}\right)$
Divide by $3^{2 \mathrm{r}}$
$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{\left(\frac{2}{3}\right)^{2 r} \cdot 2+3}\right)$
$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{3\left(\left(\frac{2}{3}\right)^{2 r+1}+1\right)}\right)$
Let $\left(\frac{2}{3}\right)^{\mathrm{r}}=\mathrm{t}$
$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3} t^{2}}\right)$
$\sum_{\mathrm{r}=1}^{\mathrm{k}}\left(\tan ^{-1}(\mathrm{t})-\tan ^{-1}\left(\frac{2 \mathrm{t}}{3}\right)\right)$
$\sum_{r=1}^{k}\left(\tan ^{-1}\left(\frac{2}{3}\right)^{r}-\tan ^{-1}\left(\frac{2}{3}\right)^{r+1}\right)$
$S_{k}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{k+1}$
$S_{\infty}=\lim _{k \rightarrow \infty}\left(\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{k+1}\right)$
$=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}(0)$
$\therefore \quad \mathrm{S}_{\infty}=\tan ^{-1}\left(\frac{2}{3}\right)=\cot ^{-1}\left(\frac{3}{2}\right)$
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