<s>[INST] How do we find the value of $\tan\biggr(\sum_{1\leq i\leq 21}f(i)\biggr)$ if $f(x) = \arctan\left(\frac{1}{x^{2 }+ x + 1}\right)$? Let's assume that $f : \mathbb{R}\rightarrow (-\frac{\pi}{2}, \frac{\pi}{2})$ such that for every $x\in \mathbb{R}$, $f(x) = \arctan\biggr(\frac{1}{x^2+x+1}\biggr)$. How can we find $\tan\biggr(\sum_{1\leq i\leq 21}f(i)\biggr)$?
Below is a fact that I've derived:
$$\tan\biggr(f(x)+f(x+1)\biggr) = \frac{\frac{1}{(x+1)^2+(x+1)+1}+\frac{1}{x^2+x+1}}{1-\frac{1}{((x+1)^2+(x+1)+1)(x^2+x+1)}} = \frac{2}{(x+1)^2}$$
But I am not sure how this can be used to approach the problem.
 [/INST] As @TheBestMagician has mentioned, notice that:
\begin{align*}
\arctan\left(\frac{1}{x^{2} + x + 1}\right) = \arctan\left(\frac{(x + 1) - x}{1 + (x + 1)x}\right) = \arctan(x + 1) - \arctan(x)
\end{align*}
Based on such identity, the proposed sum telescopes.
Hopefully this helps!
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