Extended Triangle-ception

The general formula for the total number of triangles as a function of $n$ is

$f(n) = \lfloor \dfrac{n(n + 2)(2n + 1)}{8} \rfloor,$ as found here .

The proof of this formula is quite involved. A discussion on this general problem, with a variety of approaches, can be found here .

To solve for $n$ in this problem, I first made an initial estimate using the approximation

$2n^{3} = 8*988 \Longrightarrow n = 15.81$ to 2 decimal places.

I knew this estimate would be on the high side, so I tried $f(15)$ to find that

$f(15) = \lfloor \dfrac{7905}{8} \rfloor = \lfloor 988.125 \rfloor = 988,$ and thus concluded that $n = \boxed{15}.$

I propose a programmation solution in python 3.4:

We have $n^2 <= 988$ then $n <= 31$ .

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def nbtriangles(n):
    cnt = 0
    for k in range(1, n + 1):
        for i in range(1, n + 2 -k):
            cnt += i
        for i in range(1, n + 2 -2*k):
            cnt += i
    return cnt

for n in range(1,32):
    if nbtriangles(n) == 988:
        print(n)