# Patterns

Evaluating terms of a sequence, given the formula, isn't so bad. Going the other way around is a little trickier. It can be a bit like juggling buzzing chainsaws while riding a unicycle and chewing gum.

Okay, it's not that hard. Given the terms, how do you figure out the formula?

### Sample Problem

Write down the formula for the general term in the sequence

2,4,6,8,10,...

starting with *n* = 1.

Answer.

Let's make a table so we can see the relationship between *n* and *a _{n}*.

The second number in each row is obtained by doubling the first number.

The formula for the general term is

*a _{n}* = 2

*n*.

You should be very familiar with the following common sequences and the definitions of their general terms.

- The sequence of natural numbers
*a*=_{n}*n*:

1, 2, 3, 4,...

- The sequence of squares
*a*=_{n}*n*^{2}:

1, 4, 9, 16, 25,...

- The sequence of cubes
*a*=_{n}*n*^{3}:

1, 8, 27, 64,...

- The sequence of even numbers
*a*= 2_{n}*n*:

2, 4, 6, 8,...

- The sequence of odd numbers
*a*= 2_{n}*n*– 1:

1, 3, 5, 7,...

We can also write this sequence as*a*= 2_{n}*n*+ 1 if we start with*n*= 0 instead of*n*= 1.

- The sequence of powers of 2 where
*a*= 2_{n}:^{n}

2, 4, 8, 16, 32,...

Most people don't want to reinvent the wheel, and mathematicians are no exception. Many sequences are built by making slight adjustments to more familiar sequences. Adjustments might include adding, subtracting, or multiplying by a constant.

In most of the sequences we've looked at so far, *n* only shows up once. There's only one occurrence of *n* in the formula *a _{n}* =

*n*

^{2}or

*a*= 2

_{n}*n*.

The most complicated general term we've seen is *a _{n}* =

*n*

^{2}+

*n*.

When terms get more complicated, finding a formula for the *n*^{th} term can feel like trying to solve a Rubik's cube. Just like the Rubik's cube, we have to look at individual parts to figure out the general formula.

Here's a useful trick. When you get to Taylor series, you'll need to be comfortable with sequences whose terms have alternating signs.

### Sample Problem

Find a formula for the general term of the sequence

-1, 1, -1, 1,...

starting at *n* = 1.

Answer.

If we raise (-1) to an odd power, we get -1. If we raise (-1) to an even power, we get + 1. We could think of the sequence

-1, 1, -1, 1,...

as

(-1)^{1}, (-1)^{2}, (-1)^{3}, (-1)^{4}, ...

Then the formula for the general term is

*a _{n}* = (-1)

*.*

^{n}### Sample Problem

Find a formula for the general term of the sequence

1, -1, 1, -1,...

starting at *n* = 1.

Answer.

There are (at least) two answers. We can think of this sequence as

(-1)^{2},(-1)^{3},(-1)^{4},(-1)^{5},...

in which case the general term is

*a _{n}* = (-1)

^{n + 1}.

Or we can think of the sequence as

(-1)^{0},(-1)^{1},(-1)^{2},(-1)^{3},...

in which case the general term is

*a _{n} *= (-1)

^{n – 1}.

Either formula is right.

If the terms of a sequence have alternating signs, the formula for the general term will have a factor of (-1) raised to some power.

Using this trick is simple. To find the formula for such a sequence, first ignore the alternating signs and find the formula you would have if all terms were positive. Then multiply by (-1)* ^{n}* or (-1)

^{n + 1}to account for the signs.

Next, we will teach you how to amaze your friends by levitating six inches off the floor.