Figuring out the pattern

As shown on Fig 4 below, the following transformations can be completed on any even figure and still preserve the percentage of gray triangles to all triangles. (The white triangles were changed to blue triangles to distinguish them from the white background.)

From $1$ to $2$ , swap each blue right-side up triangle with its corresponding upside down gray triangle.

From $2$ to $3$ , rotate the set of blue triangles $180°$ .

From $3$ to $4$ , double each triangle by expanding it to the blank triangle to their immediate right.

From $4$ to $5$ , skew each parallelogram to make them squares.

From $5$ to $6$ , replace each blue square under the halfway line with gray squares near the top.

In general, for each even Fig $n$ , there are $\frac{1}{2}n^2 + \frac{1}{2}n$ gray squares and a total of $n^2$ squares, for a ratio of $\frac{n^2 + n}{2n^2}$ .

Therefore, Fig $250$ will be $\frac{250^2 + 250}{2 \cdot 250^2} \cdot 100 = \boxed{50.2\%}$ gray.

The number of all triangles in Fig $n$ is given by:

$S_t (n) = \sum_{k=1}^n (2k-1) = 2\cdot \frac {n(n+1)}2 - n = n^2$

The number of grey triangles in FIg $n$ , where $n$ is even, is given by:

$S_g (n) = \sum_{k=1}^\frac n2 (4k-1) = \frac 42\left(\frac n2 \left(\frac n2 + 1\right)\right) - n = \frac {n(n+1)}2$

$\begin{aligned} \implies \frac {S_g(n)}{S_t(n)} & = \frac {n(n+1)}{2n^2} = \frac {n+1}{2n} \\ \frac {S_g(250)}{S_t(250)} & = \frac {251}{500} = \boxed{50.2} \% \end{aligned}$

Fig. 2 has 3 gray triangles on 2² = 4

Fig. 4 has 3 + 3 + 4 = 3 + 7 = 10 gray triangles on 4² = 16

Fig. 6 would have 3 + 7 + 7 + 4 = 3 + 7 + 11 = 21 gray triangles on 6² = 36

...

It seem that if fig. n (n is even) has $\Large{ \frac{n^2} {2} +\frac{n} {2}}$ gray triangles on $n^2$ or in percentage $50{\Large \left( 1 + \frac {1} {n}\right)}\%$ than in fig. (n+2) there are $\Large{ \frac{n^2} {2} +\frac{n} {2}} + 2n + 3 = \frac{1} {2} \left(n^2 + 5n + 6 \right) = \frac{\left(n + 2 \right)^2} {2} + \frac{\left(n + 2 \right)} {2}$ gray triangles on $(n + 2)^2$ or in percentage $50{\Large \left( 1 + \frac {1} {n+2}\right)}$

So it's demonstrated by induction that in fig. n (n even) there are $\Large{ \frac{n^2} {2} +\frac{n} {2}}$ gray triangles on n² or in percentage $50{\Large \left( 1 + \frac {1} {n}\right)}\%$

In particular for n= 250 percentage is $\Large {50 \left( 1 + \frac {1} {250}\right)} = 50,2\%$