I have a big triangle made up of several small triangle as depicted in picture given below.
Suppose, there is one generic triangle of this shape which is formed by joining points arranged in n rows. For example, shape given in above picture is obtained by joining points arranged in 5 rows.
I want to find sum of area of all triangles present in this picture, infact in a generic picture of same shape.
Manually, I can find it by summing area of unit triangles, area of triangles formed by 4 unit triangles, area of triangles formed by 9 unit triangles ... area of triangles formed by (n-1)^2 unit triangles ..etc.
How can I derive a generic formula to solve this problem?
The lower left unit triangle of a triangle with sides parallel to the enclosing triangles and with side length $k$ can lie anywhere in the lower left triangle of side length $n-k$, so there are $\frac{(n-k)(n-k+1)}2$ such triangles.
The top side of a triangle with the opposite orientation and with side length $k$ can be anywhere from $k$ to $n-1-k$ rows below the upper tip of the enclosing triangle, and if it's $l$ rows below the tip it has $l-k+1$ options for its horizontal position, for a total of
$$ \sum_{l=k}^{n-1-k}(l-k+1)=\frac{(n+1-2k)(n-2k)}2\;. $$
A triangle with side length $k$ contains $k^2$ unit triangles. Thus the total area in units of unit triangles is (with $r=\left\lfloor\frac{n-1}2\right\rfloor$)
$$ \sum_{k=1}^{n-1}\frac{(n-k)(n-k+1)}2k^2+\sum_{k=1}^r\frac{(n+1-2k)(n-2k)}2k^2\\ =\frac1{120}(n-1)n(2n^3+7n^2+7n+2)\\+\frac1{60}r(r+1)(24r^3-3(10n-7)r^2+(10n^2-20n-11)r+5n^2+5n-4)\;. $$