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A triangular grid is obtained by tiling an equilateral triangle of side length $n$ by $n^2$ equilateral triangles of side length $1$. Determine the number of parallelograms bounded by line segments of the grid.

There are $3$ sets of parallel lines that are being used to make each of the parallelograms. That means, if the triangle is $\triangle ABC$, then the lines that are drawn are parallel to either $AB$ or $BC$ or $CA$. Each parallelograms contains only $2$ sets of parallel lines. For symmetry, we can just ignore one set of parallel lines and just count the number of parallelograms formed by the other two, then multiply this result by $3$ to get the actual answer.

We can easily count the number of parallelograms of all sizes $1\times1,\dots, n\times 1$ and add them.
For example, the number of parallelograms of dimensions $1\times1$ is $1+2+\cdots+(n-1)=\frac12n(n-1)$
Similarly, we can find the number of parallelograms of other dimensions. They all are some easy triangle number.
Summing all of them and then multiplying by $3$, we get the following which is the number of total parallelograms:
$\frac18n(n-1)(n^2+3n+2)$

I apologise for missing the calculation details, in case you need, don't hesitate to let me know...