1.2 Exploration with Patterns

The pattern repeats every three terms: one square followed by two triangles. So the next three terms continue that cycle:

$$\square,\ \triangle,\ \triangle$$

The sequence is:

$$\square,\ \triangle,\ \triangle,\ \square,\ \triangle,\ \triangle,\ \square,\ \triangle,\ \triangle,\ \square,\ \ldots$$

It looks like the result is always a string of 1s, and the coefficient of 9 grows by appending the next digit each time. It's a satisfying pattern, but patterns can fool you.

Check what happens much further in the sequence:

$$13 + 123456789101112(9) = 1111111101910021$$

That last result is not a string of 1s. The pattern breaks down because the digits in the large multiplier of 9 start interfering with each other when they carry over. Always be suspicious of a pattern based on only a few examples.

An inductive observation is: the sum of three consecutive integers always equals three times the middle number.

• $3 + 4 + 5 = 12 = 3 \times 4$
• $14 + 15 + 16 = 45 = 3 \times 15$
• $19 + 20 + 21 = 60 = 3 \times 20$
• $99 + 100 + 101 = 300 = 3 \times 100$

We could even check it with a few more cases. Pick a random number, say $5000$. $3\times 5000 = 15000$, so the sum of the consecutive integers around it should be the same. And indeed, they are: $4999+5000+5001=15000$.

That's inductive reasoning. We noticed it from examples. To make it deductive, we have to work from rules and axioms that we know to be true. A very powerful thing you can do in mathematics is to represent values that can change with a variable. If we want to talk about 3 consecutive integers in general we could let the them be $n-1$, $n$, and $n+1$. If you pick any value for $n$, then we're talking about the 3 integers around that $n$. Next, lets try to add them together:

$$(n-1) + n + (n+1) = 3n$$

No matter what integer $n$ is, the sum is always $3n$, i.e., three times the middle number. This is now a proof. It works for every set of three consecutive integers, not just the ones we checked.

The early evidence strongly supports Amy's conjecture. Every result so far is prime. That's inductive reasoning: it's persuasive, but it's not proof.

Now try to break it. Plug in $x = 11$:

$$y = 11^2 + 11 + 11 = 121 + 11 + 11 = 143$$

Is 143 prime? Factor it: $143 = 11 \times 13$. That's two factors other than 1 and 143 itself, so 143 is not prime. Amy's conjecture is false.

One counterexample is all it takes to disprove an inductive claim.

First, identify $a_1$ and $d$.

The first term is $a_1 = 3$. Each step increases by 4, so $d = 4$.

Plug into the formula:

$$a_n = a_1 + d(n-1) = 3 + 4(n-1) = 3 + 4n - 4 = 4n - 1$$

For the 150th term, substitute $n = 150$:

$$a_{150} = 4(150) - 1 = 600 - 1 = 599$$

Start by just collecting data. Figure 1 is a lone octagon, so 8 stides. For Figure 2, the new octagon shares exactly one side with the previous one, so instead of adding a full 8 sides, you only add 7 (the shared side is already there). The same is true every time you attach the next octagon.

This is an arithmetic sequence with first term $a_1 = 8$ and common difference $d = 7$:

$$a_n = 8 + 7(n - 1) = 7n + 1$$

For the 10th figure:

$$a_{10} = 7(10) + 1 = \mathbf{71} \text{ sides}$$

Compute the first four terms for each and check whether the differences are constant.

1. $a_n = 4n + 3$

$$a_1 = 7,\quad a_2 = 11,\quad a_3 = 15,\quad a_4 = 19$$

Differences: $4, 4, 4$ which means the sequence is arithmetic with $d = 4$.

2. $a_n = n^2 - 1$

$$a_1 = 0,\quad a_2 = 3,\quad a_3 = 8,\quad a_4 = 15$$

Differences: $3, 5, 7$ which means the differences are not constant. Not arithmetic.

3. $a_n = \frac{n(n+1)}{2}$ (the triangular numbers!)

$$a_1 = 1,\quad a_2 = 3,\quad a_3 = 6,\quad a_4 = 10$$

Differences: $2, 3, 4$; not constant. Not arithmetic.

This is an arithmetic sequence with $a_1 = 20$ and $d = 4$. We want to find $n$ such that $a_n = 144$.

$$a_n = 20 + 4(n-1) = 144$$

Expand and solve:

$$20 + 4n - 4 = 144$$ $$16 + 4n = 144$$ $$4n = 128$$ $$n = 32$$

The problem says the last row has 144 seats, and we found $n = 32$. Let's verify: $a_{32} = 20 + 4(31) = 20 + 124 = 144$.

There are 32 rows in the theater.

Build the sequence term by term. Start with $f_1 = 1$, $f_2 = 1$, and apply $f_n = f_{n-2} + f_{n-1}$:

So $f_{10} = \mathbf{55}$.

Continuing:

So $f_{20} = \mathbf{6765}$.

For each, divide any term by the one before it to find $r$, then build the formula $g_n = g_1 \cdot r^{n-1}$.

1. $r = 2/1 = 2$, so $g_n = 1 \cdot 2^{n-1} = 2^{n-1}$.

2. $r = 30/3 = 10$, so $g_n = 3 \cdot 10^{n-1}$.

3. $r = 0.9/9 = 0.1 = \frac{1}{10}$, so $g_n = 9 \cdot \left(\frac{1}{10}\right)^{n-1}$.

4. $r = \frac{1/25}{1/5} = \frac{1}{5}$, so $g_n = \frac{1}{5} \cdot \left(\frac{1}{5}\right)^{n-1} = \left(\frac{1}{5}\right)^n$.

Let's build a table. The $n$th rectangular number is $n(n+1)$:

Looking down the columns, one big pattern jumps out immediately: every rectangular number is exactly twice the corresponding triangular number. That's not a coincidence. There's a beautiful geometric reason for it.

If you have $t_n$ dots arranged in a triangle, you can take another triangle of the same size, flip it upside down, and fit it perfectly next to the first one to make a rectangle. For example, $t_3 = 6$ dots form a triangle:

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Flip another copy and attach it to make a $3 \times 4$ rectangle:

  • • •      •
  • •  <-  • •
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... and you've got $3 \times 4 = 12$ dots. Since we used two triangles of size $t_3 = 6$, we have $2 \times 6 = 12$. Perfect.

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So our conjecture is:

$$\text{rectangular}_n = 2 \cdot t_n = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$$

Recognizing and working with patterns like these is a difficult skill to build. Practice seeing the beauty in mathematics.

For now it is enough to simply look for patterns here. But for those curious enough to see how a formula might be derived, let's compute the first several pentagonal numbers and build a table. Looking at the figures, you can see they are: $1, 5, 12, 22, 35, 51, 70, \ldots$

This is interesting! The first differences are $4, 7, 10, 13, 16, 19, \ldots$, and those are increasing by 3 each time. Noteably, the second differences are constant at 3.

When the second differences are constant, that tells us the sequence comes from a formula involving $n^2$. Let's guess the form is $P_n = an^2 + bn + c$ for some constants $a$, $b$, and $c$.

Using our data:

• $P_1 = a(1) + b(1) + c = 1$, so $a + b + c = 1$
• $P_2 = a(4) + b(2) + c = 5$, so $4a + 2b + c = 5$
• $P_3 = a(9) + b(3) + c = 12$, so $9a + 3b + c = 12$

Subtracting the first equation from the second: $3a + b = 4$.

Subtracting the second from the third: $5a + b = 7$.

Subtracting these two new equations: $2a = 3$, so $a = \frac{3}{2}$.

Plugging back into $3a + b = 4$: $3 \cdot \frac{3}{2} + b = 4$, so $\frac{9}{2} + b = 4$, giving $b = -\frac{1}{2}$.

Plugging into $a + b + c = 1$: $\frac{3}{2} - \frac{1}{2} + c = 1$, so $1 + c = 1$, giving $c = 0$.

Therefore:

$$P_n = \frac{3}{2}n^2 - \frac{1}{2}n = \frac{n(3n-1)}{2}$$

Let's verify: $P_3 = \frac{3(3 \cdot 3 - 1)}{2} = \frac{3(8)}{2} = 12$. $P_5 = \frac{5(15-1)}{2} = \frac{5 \cdot 14}{2} = 35$.

Notice something cool: the pentagonal formula $P_n = \frac{n(3n-1)}{2}$ looks a lot like the triangular formula $t_n = \frac{n(n+1)}{2}$. They're built the same way, just with different multipliers inside. This shows how related all these figurate number sequences are. They all follow quadratic patterns, and understanding the differences between them teaches us about families of sequences.