Can you solve the angles

First let's solve $\angle BA_2A_1 + \angle BA_3A_2 + \angle BA_4A_3$ .

We know $\angle BA_2A_1 = 45^\circ$ , since $\angle BA_1A_2 = 90^\circ$ and $\overline{A_1B} = \overline{A_1A_2} = 1$ .

In $\triangle BA_2A_3$ and $\triangle BA_4A_2 :$

$\overline{BA_2} = \sqrt{2}$ , $\overline{A_2A_4} = 2$

$\because$ $\frac{\overline{A_2A_3}}{\overline{BA_2}} = \frac{\overline{BA_2}}{\overline{A_2A_4}} = \frac{1}{\sqrt{2}}$

$\angle BA_2A_3 = \angle A_4A_2B$

$\therefore$ $\triangle BA_2A_3$ $\sim$ $\triangle A_4BA_2$ ( $SAS$ )

$\therefore$ $\angle BA_4A_3 = \angle A_3BA_2$

$\therefore$ $\angle BA_3A_2 + \angle BA_4A_3 = \angle BA_3A_2 + \angle A_3BA_2 = \angle BA_1A_2 = 45^\circ$

What pattern are we trying to prove?

You'll find that $\triangle BA_2C_n \sim \triangle A_{n+4}A_2B$ and here's why:

In $\triangle BA_2C_n$ and $\triangle BA_2A_{n+4} :$

$\overline{A_2C_n} = \overline{A_2A_3} - \overline{A_3C_n} = 1 - \frac{n}{n+2} = \frac{2}{n+2}$

$\overline{A_2A_{n+4}} = n + 4 - 2 = n + 2$

$\because$ $\overline{A_2C_n} \cdot \overline{A_2A_{n+4}} = \frac{2}{n+2} \cdot (n+2) = 2 = \sqrt{2} \cdot \sqrt{2} = \overline{BA_2}^2$

$\angle BA_2C_n = \angle A_{n+4}A_2B$

$\therefore$ $\frac{\overline{A_2C_n}}{\overline{BA_2}} = \frac{\overline{BA_2}}{\overline{A_2A_n+4}}$

$\therefore$ $\triangle BA_2C_n \sim \triangle A_{n+4}A_2B$

$\therefore$ $\angle BA_{n+4}A_2 = \angle C_nBA_2$

$\therefore$ $\angle BC_nA_2 + \angle BA_{n+4}A_2 = \angle BC_nA_2 + C_nBA_2 = \angle BA_2A_1 = 45^\circ$

You'll notice that this actually is what the question is constructed of:

We have already proven that $\angle BA_2A_1 + \angle BA_3A_2 + \angle BA_4A_3 = 90^\circ$

We also have the formula $\angle BC_nA_2 + \angle BA_{n+4}A_2 = 45^\circ$

The rest can be rearranged to get $(\angle BC_1A_2 + \angle BA_5A_2) + ...... + (\angle BC_{2016}A_2 + \angle BA_{2020}A_2) = (45 \times 2016)^\circ = 90720^\circ$ .

Then adding the previous $90^\circ$ we get to our final answer:

$90720^\circ + 90^\circ = \boxed{90810^\circ}$