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The function $f(x)=\tan^{-1}\left(\frac{1}{x^2+x+1}\right)$ and the sum $$A = \sum_{k=1}^{17}f(k)$$ is given. Evaluate $\tan(A)$.

I tried to expand the expression using trigonometric addition formula but i couldn't find any pattern or it doesn't become simple. Thanks in advance!

mathreadler
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2 Answers2

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Well, we have (try to prove that):

$$\mathscr{A}\left(\text{n}\right):=\sum_{\text{k}=1}^\text{n}\arctan\left(\frac{1}{\text{k}^2+\text{k}+1}\right)=\arctan\left(1+\text{n}\right)-\frac{\pi}{4}\tag1$$

So:

$$\tan\left(\mathscr{A}\left(\text{n}\right)\right)=\tan\left(\arctan\left(1+\text{n}\right)-\frac{\pi}{4}\right)=\frac{\text{n}}{2+\text{n}}\tag2$$

Jan Eerland
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  • Of course, this is useful only in as much as the OP is able to prove (1) on their own. And obviously they are not since if they were they would not ask the question. @upvoters Please explain how this post can be useful to anybody asking the question. – Did Apr 23 '17 at 08:26
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Hint...try simplifying $$\arctan\frac 1k-\arctan \frac{1}{k+1}$$

You will find you have a telescoping sum.

David Quinn
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