Basic Rules and Notation

Basic Nules and Rotation…We Mean Rules and Notation

First things first. Before we go writing rules for sequences willy-nilly, we need to establish a common notation, or way of writing these things down. It is handy to know that we'll typically start with the first term, which we'll call a₁. The next term will be called a₂ and then, you guessed it, a₃. We can have any number of terms.

Eventually, somewhere way down the number line, we will get to some random term we'll call the n^th term, "n" being the shorthand way of writing "any number." We call this term a_n. Go figure. We can then put a_n in braces, like so: {a_n}. This tells the reader that we're now talking about the rule that will give us the n^th term. In other words, if we put 3 in for n in our rule, we would be looking at the third term in the sequence. Let's try a few.

Sample Problem

Find the rule and a₆ for the sequence: 4, 7, 10, 13, …

There is no 100%, no-doubt, sure-fire, going-to-work-every-time method to finding the rule for a sequence. Sorry. You just need to be a good problem solver and apply what you know about different operations to the numbers you're given.

Here, the first thing we might notice is that the numbers go up by 3 every time. That means that 13 + 3 = 16 and then 16 + 3 = 19. So a₆ = 19. That's our sixth term.

For now, we want to put together a general rule that gives us any term. We need something that looks like {a_n} = ________ and uses 'n' in the actual rule, also known as an explicit rule. We're guessing we'll also need to use the number 3 somehow. Adding 3 might work, but then we have to use previous terms and we're not quite there yet. That's called a recursive rule…we'll get there later.

So what if we relied on the challenging idea of multiplication instead? Our first term (meaning n = 1) is really 1 + 3. The second is 1 + 3 + 3. The third is 1 + 3 + 3 + 3. We can conclude that the term number is the same as the number of 3s we used. That means our rule would be:

{a_n} = 1 + 3n

We like these rules because we can always check and make sure they work. For instance, if we plug in 6 for n to get the sixth term, we get:

a₆ = 1 + 3(6) = 19

Done and done.

Sample Problem

Find the rule and a₁₀ for the sequence 2, 4, 8, 16, …

This one forces us to step up our game a little bit. We don't add the same number over and over this time. Instead, we add 2, then 4, then 8. But wait; wouldn't you agree that each number is just multiplied by another 2? In other words, the first term is 2, and then the second is 2 × 2, and then 2 × 2 × 2, and then 2 × 2 × 2 × 2, and then and then and then…this calls for an exponent. Therefore, our rule could be:

{a_n} = 2ⁿ

Let's check it. If we go back and put in 3 for n, we get:

a₃ = 2³ = 8

Nice. That's our third term all right, so our rule works.

We can then go back and use our rule to get a₁₀.

a₁₀ = 2¹⁰ = 1024

Sample Problem

Find the rule and a₇ for the sequence 1, 2, 6, 24, 120, …

This problem is tough. No doubt about it. We made it that way on purpose. Shmoop is actually a league of evil educators whose sole purpose is to make your life less fun. True story.

Regardless, after taking some time to analyze the problem, you might be able to see that we multiply 1 × 2 to get 2, 2 × 3 to get 6, 2 × 3 × 4 to get 24, and so forth. So we have this idea that we need to multiply by the next biggest number in order to jump from one term to the next. This is a pretty unique math concept. In fact, we've seen it before.

It's called a factorial, and they come up a lot when talking about sequences. Math freakazoids love this stuff. Factorials are so cool, they are written with a '!', literally:

3! = 3 × 2 × 1 = 6

5! = 5 × 4 × 3 × 2 × 1 = 120.

1! is the boring number 1,  but 0! is also the number 1. Why? Ask Algebra 1.

Back to the problem at hand, where we still need a stinkin' rule for 1, 2, 6, 24, 120, …. It appears as though the first term is 1!, the second is 2!, the third is 3!, and so on. How about we make our rule for a_n:

{a_n} = n!

You can check it and see that it works with 4! and 5! as well and can now help us to find a₇. This is going to be nothing more (or less) than a₇ = 7! = 5040.