Q116 / 180 · KCET 2009 · Mathematics
$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots$ upto $\infty$ is equal to
- Correct answerA$\frac{\pi}{4}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{2}$
- D$\frac{\pi}{5}$
Solution
$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots \infty$
$=\sum_{\mathrm{r}=1}^{\infty} \cot ^{-1}\left(2 \cdot \mathrm{r}^{2}\right)=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^{2}}\right)$
$=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{(1+2 \mathrm{r})+(1-2 \mathrm{r})}{1-(1+2 \mathrm{r})(1-2 \mathrm{r})}\right)$
$=\sum_{\mathrm{r}=1}^{\infty}\left[\tan ^{-1}(1+2 \mathrm{r})+\tan ^{-1}(1-2 \mathrm{r})\right]$
$=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3+\tan ^{-1} 7-\tan ^{-1} 5+\ldots+\tan ^{-1} \infty$
$=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$
Hence, the correct answer is option: 'A'.