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Prove that $$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$$

I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}=\tan^{-1}\frac{1}{2}$ and so on but cannot observe any pattern. Could someone suggest something?

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Hint: Observe last term of $L.H.S.$ carefully and note that $\tan^{-1}a-\tan^{-1} b=\tan^{-1}\frac{a-b}{1+ab}$

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HINT: $$\sum_{r=1}^n \tan^{-1}\frac{1}{r^2+r+1}$$ $$=\sum_{r=1}^n \tan^{-1}\frac{(r+1)-r}{1+r(r+1)}$$ $$=\sum_{r=1}^n [\tan^{-1} (r+1) -\tan^{-1} r]$$

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    $\begingroup$ Well done! +1... $\endgroup$ – Mark Viola Apr 22 '16 at 16:31
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Hint: $$\arctan\frac {1}{1+k+k^2}=\arctan\frac{(k+1)-k}{1+(k+1)k}=\arctan(k+1)-\arctan k$$

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  • $\begingroup$ Well done! +1... $\endgroup$ – Mark Viola Apr 22 '16 at 16:32

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