Hexagons in n-Row Grid: Triangular Number Formula

The image provided displays three tables analyzing how many hexagons of different sizes (side length $a$) can be formed within a larger hexagonal grid of $n$ rows. The investigation reveals that the number of hexagons follows a pattern based on Triangular Numbers ($T_x$).

Answer

The total number of hexagons $H(n)$ for a grid with $n$ rows is the sum of triangular numbers $T_x$ whose indices decrease by 3 for each increase in side length $a$. The generalized formula for the total number of hexagons is $H(n) = \sum_{k=0}^{\lfloor (n-3)/3 \rfloor} T_{n-2-3k} = \sum_{k=0}^{\lfloor (n-3)/3 \rfloor} \frac{(n-2-3k)(n-1-3k)}{2}$.

Explanation

1. Observation of the Pattern in Table 1 In Table 1, we observe that for a hexagon of side length $a=1$, the number of hexagons possible follows the sequence of triangular numbers starting from $n=3$.

   • For $n=3, a=1$, count = $1 (T_1)$.
   • For $n=4, a=1$, count = $3 (T_2)$.
   • For $n=5, a=1$, count = $6 (T_3)$. The formula for the first column is $T_{n-2}$, where $T_x = \frac{x(x+1)}{2}$.

2. Identifying the Threshold for Larger Hexagons Table 2 shows that a hexagon of side length $a=2$ only becomes possible when $n=6$. A side length $a=3$ becomes possible at $n=9$. ⚠️ This step is required on exams: Notice that the side length $a$ can only increase every 3 rows. This implies the index of the triangular number for the next column is shifted by 3.

   • Column $a=1$ uses $T_{n-2}$.
   • Column $a=2$ uses $T_{n-5}$.
   • Column $a=3$ uses $T_{n-8}$.

3. Generalizing the Column Entry Any entry for a grid of size $n$ and hexagon of side $a$ can be written as: $T_{n - (3a - 1)}$ This formula gives the number of hexagons of side length $a$ that fit in $n$ rows.

4. Summing for the Total Count Table 3 calculates the "Total" by adding the values in each row. To find the total hexagons in a grid of $n$ rows, we sum the possible $a$ values until $3a-1 > n$. $H(n) = T_{n-2} + T_{n-5} + T_{n-8} + \dots$ Using the sum of triangular numbers often leads to a cubic relationship, but because the terms skip by 3, we treat it as a finite series.

5. Linking to the Goal Formula The text mentions the sum of squares formula $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$. This is relevant because triangular numbers are quadratic, and the sum of triangular numbers involves $n^3$ terms. This level of sophistication is used to simplify the series into a single polynomial expression.

Final Answer

The total number of hexagons $H$ for a grid of $n$ rows is given by the sum of triangular numbers with indices decreasing by 3: $\boxed{H(n) = \sum_{k=0}^{\lfloor \frac{n-3}{3} \rfloor} \frac{(n-2-3k)(n-1-3k)}{2}}$

Common Mistakes

• Incorrect Indexing: Starting the triangular number sequence at $T_n$ instead of $T_{n-2}$. Always check the $n=3$ case; if you use $T_n$, you would get 6 instead of 1.
• Missing the "Every 3rd Row" Rule: Students often assume a new hexagon size $a$ appears for every new row $n$. The geometry of hexagons requires 3 additional rows to increase the side length $a$ by 1 unit.