To solve an arctan sum $\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)}$ I was suggested (without any background setting) to prove a numerically checked result, which is dramatically refined
$$
\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)} = \frac{\pi}{12}
$$
where $a_n=(1+\sqrt3)^{4^n}$, with $4^n$ on shoulder.
Usually we deal with arctan series using telescoping or gamma function technique, but this $a_n$ holds a geometrical growth and I have no idea how to do the reduction. Thanks in advance for any suggestion.

Let $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. Since $a_{n+1} = a_n^4$, we have
$b_{n+1} = \frac{\sqrt{3}}{1 + 2a_{n+1}} = \frac{\sqrt{3}}{1 + 2a_n^4}$.
With help of a CAS, one can verify
$$\frac{\frac{\sqrt{3}}{1+2x} - \frac{\sqrt{3}}{1+2x^4}}{1 + \frac{\sqrt{3}}{1+2x}\frac{\sqrt{3}}{1+2x^4}}
= \frac{\sqrt{3}x(x-1)}{(x+1)(2x^2-3x+2)}
$$
Substitute $x$ by $a_n$, $x^4$ by $a_{n+1}$ and sum over $n$. The sum
at hand becomes
$$\begin{align}\sum_{n=0}^\infty \tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}
&= \sum_{n=0}^\infty\left(\tan^{-1}b_n - \tan^{-1}b_{n+1}\right)
= \tan^{-1}b_0\\ &= \tan^{-1}\frac{\sqrt{3}}{3+2\sqrt{3}}
= \tan^{-1}\frac{\frac12}{1+\frac{\sqrt{3}}{2}}
= \tan^{-1}\frac{\sin\frac{\pi}{6}}{1+\cos\frac{\pi}{6}}\\
&= \frac12\cdot \frac{\pi}{6} = \frac{\pi}{12}
\end{align}
$$
Update
In case anyone wonder how the magic $b_n$ is chosen. The basic idea is to rewrite the summands into the form $\tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}$ so that one can use  addition formula of tangent to turn the sum into a telescoping one. Since the summands have the form $\tan^{-1}\sqrt{3}(\cdots)$, one expect $b_n$ equals to $\sqrt{3}$ multiply something simple in $a_n$.
We know the series sum to $\tan^{-1}b_0$. Since $\tan\frac{\pi}{12} = \frac{\sqrt{3}}{3+2\sqrt{3}}$ and $a_0 = 1+\sqrt{3}$. The simplest candidate of $b_n$ comes to my mind is $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. It turns out this works.
