Dividing Diamonds

Observe that in ${n}^{\text{th}}$ case the number of quadrilaterals is given by the formula:-
$4 × (\displaystyle \sum_{i=1}^{\displaystyle \sum_{i=0}^{n-2} 2^i} i) + 4$

So, in the ${2016}^{\text{th}}$ figure there would be $4 × (\displaystyle \sum_{i=1}^{\displaystyle \sum_{i=0}^{2014} 2^i} n) + 4$

= 8384516 triangles. $_\square$

The number of triangles of n-th figure is $4(\frac{{2^{n-1}}\times{(2^{n-1}+1)}}{2})+4$

Observe that the figure is symmetric, so we can calculate in a quarter and then multiply by $4$ . Also, add $4$ to the result because there are 4 "half-square"s haven't been counted.

In a quarter of $1^\text{st}$ figure there are $2^0 = 1$ small triangle.

In a quarter of $2^\text{nd}$ figure there are $2^1 = 2$ small triangles. The number of triangles (including larger ones) in this quarter is: $2 + {2 \choose 2} = 3$ .

In a quarter of $n^\text{th}$ figure there are $2^{n-1}$ small triangles. The number of triangles (including larger ones) in this quarter is: $2^{n-1} + {2^{n-1} \choose 2}$ .

So, mulitply it by $4$ and add $4$ gives us: $2^{n+1} + 4 \times {2^{n-1} \choose 2} + 4$ .

Hence, in the $12^\text{th}$ figure, the answer should be $2^{13} + 4 \times {2^{11} \choose 2} +4 =\boxed{8392708}$ .

But the answer to this question is different. Am I wrong at somewhere?