, 2022

Answer: $\frac{n(n+1)(n+2)(n+3)(n^2+3n+6)}{240}$.

Any equilateral triangle $T$ under consideration can be embedded into an unique equilateral triangle $M$, such that the vertices of $T$ lie on the sides of $M$ and those sides are parallel to the sides of the bigger triangle. If $M$ has $k$ times bigger side than the side of a unit triangle ($k = 1, 2, \dots, n$) then there are $1 + 2 + \dots + (n+1-k) = \binom{n+2-k}{2}$ choices of $M$. We find the sum of the areas of all $T$ by enumerating the knots of one of the sides of $M$ with $0, 1, \dots, k$ and realizing that there is exactly one equilateral triangle with vertex at $i$-th knot for each $i = 1, 2, \dots, k$ and the area of this triangle is $k^2 - 3i(k-i)$. Thus the sum of the areas of all such $T$ equals
$$\sum_{i=1}^{k} (k^2 - 3i(k - i)) = k^3 - 3\binom{k+1}{3} = \frac{k^3 + k}{2}.$$
We used the well known equality $\sum_{i=1}^{k} i(k - i) = \binom{k+1}{3}$. Therefore the desired sum equals:
$$\sum_{k=1}^{n} \frac{k^3 + k}{2} \binom{n+2-k}{2} = \frac{1}{2} \sum_{k=1}^{n} k^3 \binom{n+2-k}{2} + \frac{1}{2} \sum_{k=1}^{n} k \binom{n+2-k}{2}.$$
To find the second sum count the number of words with 4 letters "a" and $n-1$ letters "b" such that the second letter "a" appears on $(k+1)$-th place. There are $k$ choices for the first "a" and $n+3-(k+1) = n+2-k$ choices for the last two letters "a". Therefore $\sum_{k=1}^{n} k \binom{n+2-k}{2} = \binom{n+3}{4}$.