Question

The sum of series $\cot ^{-1}\left(\frac{9}{2}\right)+\cot ^{-1}\left(\frac{33}{4}\right)+\cot ^{-1}\left(\frac{129}{8}\right)+\ldots \ldots \ldots . . . \infty$ is equal to

• A. $\cot ^{-1}(2)$
• B. $\cot ^{-1} 3$
• C. $\cot ^{-1}(-1)$
• D. $\cot ^{-1}(1)$

Answer 1

Answer: (A)

$S=\cot ^{-1}\left(\frac{9}{2}\right)+\cot ^{-1}\left(\frac{33}{4}\right)+\cot ^{-1}\left(\frac{129}{8}\right)+\ldots \infty$
$T _1=\cot ^{-1}\left(\frac{9}{2}\right)=\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left(\frac{2}{1+4 \times 2}\right)=\tan ^{-1} 4-\tan ^{-1} 2 $
$T _2=\cot ^{-1}\left(\frac{9}{2}\right)=\tan ^{-1}\left(\frac{4}{33}\right)=\tan ^{-1}\left(\frac{4}{1+8 \times 4}\right)=\tan ^{-1} 8-\tan ^{-1} 4$
$T _{ n }=\tan ^{-1} 2^{ n +1}-\tan ^{-1} 2^{ n } $
$S _{ n }=\Sigma T _{ n }=\tan ^{-1} 2^{ n +1}-\tan ^{-1} 2 $
$\text { as } n \rightarrow \infty$
$S _{\infty}=\frac{\pi}{2}-\tan ^{-1} 2=\cot ^{-1} 2$