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Find the sum of the first $50$ terms of the series $$\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....$$

$$ \sum_1^{50}=\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....\\ =\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}+.....= $$ My reference gives the solution $\tan^{-1}\dfrac{5}{6}$, but I do not have any clue of doing it ?

Note: I know that $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ if $xy<1$.

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    $\begingroup$ Note that $$\tan^{-1} x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$ $\endgroup$
    – Qurultay
    Dec 15, 2018 at 5:33
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    $\begingroup$ what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $\sum$ notation? $\endgroup$
    – clathratus
    Dec 15, 2018 at 5:36
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    $\begingroup$ @clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$. $\endgroup$
    – farruhota
    Dec 15, 2018 at 5:41
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    $\begingroup$ $a_n=1+n+n^2$, I guess $\endgroup$ Dec 15, 2018 at 5:45
  • $\begingroup$ Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it. $\endgroup$
    – Empy2
    Dec 15, 2018 at 5:51

1 Answer 1

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Hint

  • Use the fact that $\cot^{-1}(x)=\tan^{-1}\frac{1}{x}$

  • $\tan^{-1}\left(\frac{1}{n^2+n+1}\right) = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}(n+1) -\tan^{-1}(n)$

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