# Let Sk

Question:

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

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

2. (2) $\frac{\pi}{2}$

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

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

Correct Option: , 3

Solution:

$\mathrm{S}_{\mathrm{k}}=\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{6^{\mathrm{r}}}{2^{2 \mathrm{r}+1}+3^{2 \mathrm{r}+1}}\right)$

Divide by $3^{2 x}$

$\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{\mathrm{r}}}{\left(\frac{2}{3}\right)^{2 \mathrm{r}} \cdot 2+3}\right)$

$\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{\mathrm{r}}}{3\left(\left(\frac{2}{3}\right)^{2 \mathrm{r}+1}+1\right)}\right)$

$\operatorname{Let}\left(\frac{2}{3}\right)^{\mathrm{r}}=\mathrm{t}$

$\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{\frac{\mathrm{t}}{3}}{1+\frac{2}{3} \mathrm{t}^{2}}\right)$

$\sum_{\mathrm{r}-1}^{\mathrm{k}} \tan ^{-1}\left(\frac{\mathrm{t}-\frac{2 \mathrm{t}}{3}}{1+\mathrm{t} \cdot \frac{2 \mathrm{t}}{3}}\right)$

$\sum_{\mathrm{r}=1}^{\mathrm{k}}\left(\tan ^{-1}(\mathrm{t})-\tan ^{-1}\left(\frac{2 \mathrm{t}}{3}\right)\right)$

$\sum_{\mathrm{r}=1}^{\mathrm{k}}\left(\tan ^{-1}\left(\frac{2}{3}\right)^{\mathrm{r}}-\tan ^{-1}\left(\frac{2}{3}\right)_{\mathrm{k}+1}^{\mathrm{r}+1}\right)$

$\mathrm{S}_{\mathrm{k}}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)$

$\mathrm{S}_{\infty}=\lim _{\mathrm{k} \rightarrow \infty}\left(\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{\mathrm{k}+1}\right.$

$=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}(0)$

$\therefore \quad \mathrm{S}_{\infty}=\tan ^{-1}\left(\frac{2}{3}\right)=\cot ^{-1}\left(\frac{3}{2}\right)$

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