If $\left.\cot ^{-1} ( \alpha\right)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18$
$+\cot ^{-1} 32+\ldots .$ upto 100 terms, then $\alpha$ is :
Correct Option: 1
$\operatorname{Cot}^{-1}(\alpha)=\cot ^{-1}(2)+\cot ^{-1}(8)+\cot ^{-1}(18)+\ldots . .$
$=\sum_{\mathrm{n}=1}^{100} \tan ^{-1}\left(\frac{2}{4 \mathrm{n}^{2}}\right)$
$=\sum_{\mathrm{n}=1}^{100} \tan ^{-1}\left(\frac{(2 \mathrm{n}+1)-(2 \mathrm{n}-1)}{1+(2 \mathrm{n}+1)(2 \mathrm{n}-1)}\right)$
$=\sum_{\mathrm{n}=1}^{100} \tan ^{-1}(2 \mathrm{n}+1)-\tan ^{-1}(2 \mathrm{n}-1)$
$=\tan ^{-1} 201-\tan ^{-1} 1$
$=\tan ^{-1}\left(\frac{200}{202}\right)$
$\therefore \cot ^{-1}(\alpha)=\cot ^{-1}\left(\frac{202}{200}\right)$
$\alpha=1.01$
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