String of Squares

By Pythagorean theorem, the height from vertex $C$ to base $AB$ = $\sqrt{125^2 - 35^2} = 120$

A radius of an inscribed circle inside a triangle = $\dfrac{2\times \text{Area of triangle}}{\text{perimeter}} = \dfrac{70\times 120}{70+125+125} = \dfrac{8400}{320} = \dfrac{105}{4}$ .

Therefore, the biggest incircle's diameter = $\dfrac{105}{2} = 52.5$

Then for the next second incircle, the new height will be $120 - 52.5 = 67.5$ . Thus, the ratio of the similar smaller triangle to the original one = $\dfrac{67.5}{120} =\dfrac{9}{16}$ .

Because of similarity, the ratio for the next recursive incircle will always be $\dfrac{9}{16}$ , and so it is a geometric series for the blue squares with each square's area = $\dfrac{d^2}{2}$ where $d$ is the diameter of the incircle or the diagonal of the square. It should be noted that the ratio of the next recursive square is $(\dfrac{9}{16})^2$ as two sides are multiplied.

Thus, the total blue squares' area = $\dfrac{1}{2}(\dfrac{52.5^2}{1 - (\dfrac{9}{16})^2}) = \dfrac{52.5^2}{2}\times \dfrac{256}{175} = 2016$ .

Now don't forget the little last pink unit square. The total colored area will then equal $2016+1 = \boxed{2017}$ .