MCQ
The sum $\sum\limits_{n = 1}^\infty  {{{\cot }^{ - 1}}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to
  • $\frac{{3\pi }}{4} + {\cot ^{ - 1}}2$
  • B
    $\frac{\pi }{2} + {\cot ^{ - 1}}3$
  • C
    $\pi $
  • D
    $\frac{\pi }{2} + {\tan ^{ - 1}}2$

Answer

Correct option: A.
$\frac{{3\pi }}{4} + {\cot ^{ - 1}}2$
a
$\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{3}{\mathrm{n}^{2}+\mathrm{n}-1}\right)$

$=\tan ^{-1}(\mathrm{n}+2)-\tan ^{-1}(\mathrm{n}-1)$

Use $: \mathrm{S}_{\mathrm{n}}=\Sigma \mathrm{T}_{\mathrm{n}}$

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