How many different sets of three points in this equilateral triangular grid are the vertices of an equilateral triangle? Justify your answer.

[center][img]http://i.imgur.com/S6RXkYY.png[/img][/center]
To determine the number of different sets of three points in an equilateral triangular grid that form the vertices of an equilateral triangle, we can use a combinatorial approach. The formula provided in the solution is:

\[
\sum_{k=1}^n k \left( \frac{(n+1-k)(n+2-k)}{2} \right)
\]

This formula can be simplified to:

\[
\frac{n^4 + 6n^3 + 11n^2 + 6n}{24} = \frac{n(n+1)(n+2)(n+3)}{24}
\]

This formula counts the number of equilateral triangles that can be formed in an equilateral triangular grid of side length \( n \).

1. **Understanding the Formula:**
   - The term \( \sum_{k=1}^n k \left( \frac{(n+1-k)(n+2-k)}{2} \right) \) represents the sum of the number of triangles of different sizes that can be formed.
   - The simplification to \( \frac{n(n+1)(n+2)(n+3)}{24} \) is derived from combinatorial arguments and can be verified through algebraic manipulation.

2. **Applying the Formula:**
   - For \( n = 4 \), we substitute \( n \) into the simplified formula:
     \[
     \frac{4 \cdot 5 \cdot 6 \cdot 7}{24}
     \]

3. **Calculating the Result:**
   - First, calculate the product in the numerator:
     \[
     4 \cdot 5 = 20
     \]
     \[
     20 \cdot 6 = 120
     \]
     \[
     120 \cdot 7 = 840
     \]
   - Then, divide by the denominator:
     \[
     \frac{840}{24} = 35
     \]

Thus, the number of different sets of three points in the equilateral triangular grid that form the vertices of an equilateral triangle is \( \boxed{35} \).