$f(x) = \operatorname{arctan}\left( \frac{1}{x^2 + x + 1} \right)$ and $A = f(1) + f(2) + \cdots + f(21)$. What is the value of $\tan(A)$?
The answer is $\frac{21}{23}$, but I cant solve it. Only thing I have made is recognize that if we call $S(n) = f(1) + f(2) + \cdots + f(n)$, we see the result of $\tan(S(n)) = \frac{n}{n+2}$. So we can prove it by induction. But I wonder is there any other and smarter way to solve this problem.
Edit: For when you said $f(x)=\frac{1}{x^2+x+1}$
$$ A = f(1) + \cdots + f(21) = \frac{2549394161224968142190918333533}{3386891079486272288992946945151}\\ \tan A \approx .936697 $$
Edit: For when you put $f(x)=\tan^{-1} \frac{1}{x^2+x+1} = \tan^{-1} F(x)$
$$ A = B + f(21)\\ \tan A = \frac{\tan B + \tan f(21)}{1-\tan B \tan f(21)}\\ = \frac{\tan B + F(21)}{1-\tan B \times F(21)}\\ $$
Keep proceeding down with $B=C+f(20)$ etc.
My comment shows $\displaystyle \sum_{n=1}^{21} \arctan{\frac{1}{n^2+n+1}}=\frac{\pi}{4}-\arctan{\frac{1}{22}}$
This lead your answer.
Note that $$\tan ^{-1} (a) + \tan ^{-1} (b) = \tan ^{-1} (\frac {a+b}{1-ab})$$
We can use the above identity to compute the answer but it takes a long time.
