Find your own patterns or relations in and among the sequences in Table 1. Can you explain why they happen with a picture or otherwise?

• The terms are known as centered hexagonal numbers.
• The general term $a_n$ can be expressed in two ways: $$a_n = 3n^2 - 3n + 1$$ This was derived by finding the second differences of the sequence.
• Another pattern for the general term is that each term is the difference of two consecutive cubes: $$a_n = n^3 - (n-1)^3$$ Let's verify this: For $n=1$: $a_1 = 1^3 - (1-1)^3 = 1^3 - 0^3 = 1$. For $n=2$: $a_2 = 2^3 - (2-1)^3 = 8 - 1^3 = 7$. For $n=3$: $a_3 = 3^3 - (3-1)^3 = 27 - 2^3 = 19$. This form is equivalent to the first form: $$n^3 - (n-1)^3 = n^3 - (n^3 - 3n^2(1) + 3n(1)^2 - 1^3) = n^3 - n^3 + 3n^2 - 3n + 1 = 3n^2 - 3n + 1$$
• The sum of the first $n$ terms of this sequence is $S_n = n^3$.

The most significant explanation comes from the pattern $a_n = n^3 - (n-1)^3$.
When we sum the terms using this form, we get a telescoping sum:

Let's write out the first few terms of the sum:

Notice that most of the terms cancel out:
The $-0^3$ remains.
The $1^3$ from the first term cancels with the $-1^3$ from the second term.
The $2^3$ from the second term cancels with the $-2^3$ from the third term.
This pattern continues until the $(n-1)^3$ term.
The only terms that do not cancel are the first part of the first term ($1^3$) and the last part of the last term ($n^3$).

This explains why the sum of the first $n$ centered hexagonal numbers is always $n^3$.