## Pre-Calculus For Dummies, 2nd Edition (2012)

### Part III. Analytic Geometry and System Solving

### Chapter 14. Sequences, Series, and Expanding Binomials for the Real World

*In This Chapter*

Exploring the terms and formulas of sequences

Grasping arithmetic and geometric sequences

Summing sequences to create a series

Applying the binomial theorem to expand binomials

You can breathe a sigh of relief: In this chapter you get to put aside your graph paper and the many complex, intangible concepts that pre-calculus presents to you, such as the unit circle, conics, and logs. This chapter is dedicated to how you can use pre-calculus in the real world. The real-world applications from previous chapters are each useful to probably a handful of people. This chapter is different because the applications are useful to *everyone.* No matter who you are or what you do, you should probably understand the value of your belongings. For examples, you may want to know your car’s worth after a number of years and what your credit card or loan balance will be if you don’t pay it on time. We focus on the following topics to take math out of the classroom and into the fresh air:

**Sequences:** This application of pre-calc helps you understand patterns. You can see patterns develop, for example, in how much your car depreciates, how credit-card interest builds, and how scientists estimate the growth of bacteria populations.

**Series: **Series help you understand the sum of a sequence of numbers, such as annuities, the height a ball bounces (if you really want to figure that out in your free time), and so on.

This chapter dives into these topics and debunks the myth that math isn’t useful in the real world.

**Speaking Sequentially: Grasping the General Method**

A *sequence* is basically an ordered list of numbers following some sort of pattern. This pattern can usually be described by a general rule that allows you to find out any of the numbers in this list without having to find *all* the numbers in between. It’s infinite, meaning it can continue in the same pattern forever. A sequence’s mathematical definition is a function defined over the set of positive integers, usually written in the following form:

{*a _{n}*} =

*a*

_{1},*a*

_{2},*a*. . . ,

_{3},*a*. . .

_{n},The {*a _{n}*} portion represents the notation for the entire set of numbers. Each

*a*is called a

_{n}*term of the sequence;*

*a*is the first term,

_{1}*a*is the second term, and so on. The

_{2}*a*is the

_{n}*n*th term, meaning it can be any term you need it to be.

In the real world, sequences are helpful when describing any quantity that increases or decreases with time — financial interest, debt, sales, populations, and asset depreciation or appreciation, to name a few. All quantities that change with time based on a certain percentage follow a pattern that can be described using a sequence. Depending on the rule for a particular sequence, you can multiply the initial value of an object by a certain percentage to find a new value after a certain length of time. Repeating this process reveals the general pattern and the change in value for the object.

**Calculating a sequence’s terms by using the sequence expression**

The general formula for any sequence involves the letter *n,* which is the number of the term (the first term would be *n* = 1, and the 20th term would be *n* = 20), as well as the rule to find each term. You can find any term of a sequence by plugging *n* into the general formula, which gives you specific instructions on what to do with this value *n.* If you’re given a few terms of a sequence, you can use these terms to find the general formula for the sequence. If you’re given the general formula (complete with *n* as the variable), you can find any term by plugging in the number of the term you want for *n.*

Unless otherwise noted, the first term of any sequence {*a _{n}*} begins with

*n*= 1. The next

*n*always goes up by 1.

For example, you can use the formula to find the first three terms of *a _{n}* = (–1)

^{n}^{ – 1 }· (

*n*

^{2}):

**1.** **Find a_{1} first by plugging in 1 wherever you see n.**

*a _{1}* = (–1)

^{1 – 1 }· (1

^{2}) = (–1)

^{0}· 1 = 1 · 1 = 1

**2.** **Continue plugging in consecutive integers for n.**

This process gives you terms two and three:

*a _{2}* = (–1)

^{2 – 1}· (2

^{2}) = (–1)

^{1}· 4 = –1 · 4 = –4

*a _{3}* = (–1)

^{3 – 1}· (3

^{2}) = (–1)

^{2}· 9 = 1 · 9 = 9

**Working in reverse: Forming an expression from terms**

If you know the first few terms of a sequence, you can write a general expression for the sequence to find the *n*th term. To write the general expression, you must look for a pattern in the first few terms of the sequence, which demonstrates logical thinking (and we all want to be logical thinkers, right?). The formula you write must work for every integer value of *n,* starting with *n* = 1.

Sometimes this calculation is an easy task, and sometimes it’s less apparent and more complicated. Sequences involving fractions and/or exponents tend to be more complicated and less obvious in their patterns. The easy ones to write include addition, subtraction, multiplication, or division by integers.

For example, to find the general formula for the *n*th term of the sequence 2/3, 3/5, 4/7, 5/9, 6/11, you should look at the numerator and the denominator separately:

The numerators begin with 2 and increase by one each time. This sequence is described by *a _{n} *=

*n*+ 1.

The denominators start with 3 and increase by two each time. This sequence is described by *a _{n} *= 2

*n*+ 1.

Therefore, this sequence can be expressed by this general formula:

To double check your formula and ensure that the answers work, plug in 1, 2, 3, and so on to make sure you get the original numbers from the given sequence.

They all work, so we did it right!

**Recursive sequences: One type of general sequence**

A *recursive sequence* is a sequence in which each term depends on the term before it. To find any term in a recursive sequence, you use the given term (at least one term, usually the first, is given for the problem) and the given formula that allows you to find the other terms.

You can recognize recursive sequences because the given formula typically has *a _{n}* (the

*n*th term of the sequence) as well as

*a*(the term before the

_{n – 1}*n*th term of the sequence). In these problems, you’re given a formula (a different one for each problem), and the directions ask you to find the terms of the sequence.

For example, the most famous recursive sequence is the Fibonacci sequence, in which each term after the second term is defined as the sum of the two terms before it. The first term of this sequence is 1, and the second term is 1 also. The formula for the Fibonacci sequence is *a _{n}* =

*a*+

_{n – 2}*a*for

_{n – 1}*n*≥ 3.

So if you were asked to find the next three terms of the sequence, you’d have to use the formula as follows:

*a _{3}* =

*a*+

_{3 – 2}*a*=

_{3 – 1}*a*+

_{1}*a*= 1 + 1 = 2

_{2}*a _{4}* =

*a*+

_{4 – 2}*a*

_{4 – 1}_{ }=

*a*+

_{2}*a*= 1 + 2 = 3

_{3}*a _{5}* =

*a*+

_{5 – 2}*a*=

_{5 – 1 }*a*+

_{3}*a*= 2 + 3 = 5

_{4}The first ten terms of this sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. It is very famous because many things in the natural world follow the pattern of the Fibonacci sequence. For examples, the florets in the head of a sunflower form two oppositely directed spirals, 55 of them clockwise and 34 counterclockwise; lilies and irises both have 3 petals; buttercups have 5 petals; and corn marigolds have 13 petals. Seeds of coneflowers and sunflowers have also been observed to follow the same pattern as the Fibonacci sequence. Pine cones and cauliflower also follow this pattern.

**Covering the Distance between Terms: Arithmetic Sequences**

One of the most common types of sequences is called an *arithmetic sequence.* In an arithmetic sequence, each term differs from the one before it by the same number, called the *common difference.* To determine whether a sequence is arithmetic, you subtract each term by its preceding term; if the difference between each term is the same, the sequence is arithmetic.

Arithmetic sequences are very helpful to identify because the formula for the *n*th term of an arithmetic sequence is always the same:

*a _{n}* =

*a*+ (

_{1}*n*– 1)

*d*

where *a*_{1} is the first term and *d* is the common difference.

For exercises involving arithmetic sequences, you’re asked to find a term somewhere in a given sequence. You can recognize the sequence as arithmetic because a common difference is between each term. The general formula to find the desired terms in any arithmetic sequence has three steps: finding the common difference, writing the formula for the specific given sequence using the first term and the common difference, and then finding the term you’re asked to find by plugging in the number of the term for *n.*

In your pre-calculus course and in life, you may encounter two main types of arithmetic-sequence problems: one where you’re given a list of consecutive terms (which is easy), and one where you’re given two terms that are not consecutive (where finding the common difference is no piece of cake). In the next two sections we show you how to handle each type.

**Using consecutive terms to find another in an arithmetic sequence**

If you’re given two consecutive terms of an arithmetic sequence, the common difference between these terms is not too far away.

For example, an arithmetic sequence is –7, –4, –1, 2, 5. . . . If you want to find the 55th term of this arithmetic sequence, you can continue the pattern begun by the first few terms 50 more times. However, that process would be very time consuming and not very effective to find terms that come later in the sequence.

Instead, you can use a general formula to find any term of an arithmetic sequence. Finding the general formula for the *n*th term of an arithmetic sequence is easy as long as you know the first term and the common difference.

**1.** **Find the common difference.**

To find the common difference, simply subtract one term from the one after it: –4 – (–7) = 3.

**2.** **Plug a_{1} and d into the general formula for any arithmetic sequence to write the specific formula for the given sequence.**

Start with this equation:

*a _{n}* =

*a*+ (

_{1}*n*– 1)

*d*

Then plug in what you know: The first term of the sequence is –7, and the common difference is 3:

*a _{n}* = –7 + (

*n*– 1)3 = –7 + 3

*n*– 3 = 3

*n*– 10

**3.** **Plug in the number of the term you’re trying to find for n.**

To find the 55th term, plug 55 in for *n* into the general formula for *a _{n}:*

*a _{55}* = 3(55) + 1 = 165 + 1 = 166

**Using any two terms**

At times you’ll need to find the general formula for the *n*th term of an arithmetic sequence without knowing the first term or the common difference. In this case, you’re given two terms (not necessarily consecutive), and you use this information to find *a _{1}* and

*d.*Your steps are still the same: Find the common difference, write the specific formula for the given sequence, and then find the term you’re looking for (we can’t say that often enough).

For instance, to find the general formula of an arithmetic sequence where *a _{4}* = –23 and

*a*= 40, follow these steps:

_{22}**1.** **Find the common difference.**

You have to be more creative in finding the common difference for these types of problems.

**a.** **Use the formula a_{n} = a_{1} + (n – 1)d to set up two equations that use the given information.**

For the first equation, you know that when *n* = 4, *a _{n}* = –23:

–23 = *a _{1}* + (4 – 1)

*d*

–23 = *a _{1}* + 3

*d*

For the second equation, you know that when *n* = 22, *a _{n}* = 40:

40 = *a _{1}* + (22 – 1)

*d*

40 = *a _{1}* + 21

*d*

**b.** **Set up a system of equations (see Chapter 13) and solve for d.**

The system looks like this:

You can use elimination or substitution to solve the system, like we show you in Chapter 13. Elimination works nicely because you can multiply either equation by –1 and add the two together to get 63 = 18*d. *Therefore, *d*= 3.5.

**2.** **Write the formula for the specific sequence.**

This step is also a little more work than before.

**a.** **Plug d into one of the equations to solve for a_{1}.**

You can plug 3.5 back into either equation:

–23 = *a _{1}* + 3(3.5), or

*a*= –33.5.

_{1}**b.** **Use a_{1} and d to find the general formula for a_{n}.**

This step becomes a simple three-step simplification:

*a _{n} *= –33.5 + (

*n*– 1)3.5

*a _{n}* = –33.5 + 3.5

*n*– 3.5

*a _{n} *= 3.5

*n*– 37

**3.** **Find the term you were looking for.**

We didn’t ask in the directions to this problem to find any specific term (always read the directions!), but if we did, you could plug that number in for *n* and then find the term you were looking for.

**Sharing Ratios with Consecutive Paired Terms: Geometric Sequences**

A *geometric sequence* is one in which consecutive terms have a common ratio. In other words, if you divide each term by the term before it, the quotient, denoted by the letter *r,* is the same.

Certain objects, such as cars, depreciate with time. You can describe this depreciation by using a geometric sequence. The common ratio is always the rate as a percent (sometimes called APR, which stands for annual percentage rate). Finding the value of the car at any time, as long as you know its original value, is easy to do. The following sections show you how to identify the terms and expressions of geometric sequences, which allow you to apply the sequences to real-world situations (such as trading in your car!).

Here we begin to work with geometric sequences: how to find a term in the sequence as well as how to find the formula for the specific sequence when you’re not given it. But first, here are some general ideas to remember.

The first term of any sequence is denoted as *a _{1}.* To find the second term of a geometric sequence, multiply the first term by the common ratio,

*r.*You can follow this pattern infinitely to find any term of a geometric sequence:

{*a _{n}*} =

*a*. . . ,

_{1}, a_{2}, a_{3}, a_{4}, a_{5},*a*. . .

_{n}{*a _{n}*} =

*a*·

_{1}, a_{1}*r,*

*a*·

_{1}*r*

^{2},

*a*·

_{1}*r*

^{3},

*a*·

_{1}*r*

^{4}, . . . ,

*a*·

_{1}*r*

^{n}^{ – 1}, . . .

More simply put, the formula for the *n*th term of a geometric sequence is

*a _{n}*

_{ }=

*a*·

_{1}*r*

^{n}^{ – 1}

In the formula, *a _{1}* is the first term and

*r*is the common ratio.

**Identifying a term when you know consecutive terms**

The steps for dealing with geometric sequences are remarkably similar to those in the arithmetic sequence sections. You find the common ratio (not the difference!), you write the specific formula for the given sequence, and then you find the term you’re looking for.

An example of a geometric sequence is 2, 4, 8, 16, 32. To find the 15th term, follow these steps:

**1.** **Find the common ratio.**

In this sequence, each consecutive term is twice the previous term. If you can’t see the common difference by looking at the sequence, divide any term by the term before it.

**2.** **Find the formula for the given sequence.**

In terms of the formula, *a _{1}* = 2 and

*r*= 2. The general formula for this sequence is

*a*= 2 · 2

_{n}

^{n }^{– 1}, which simplifies (using the rules of exponents) to 2

^{1}· 2

^{n}^{ – 1}= 2

^{1 + (n – 1)}= 2

*.*

^{n}**3.** **Find the term you’re looking for.**

If *a _{n}* = 2

*, then*

^{n}*a*= 2

_{15}^{15}= 32,768.

The formula in the previous example simplifies nicely because the bases of the two exponents are the same. If the first term and *r* don’t have the same base, you can’t combine them. (For more on rules such as this, head to Chapter 5.)

**Going out of order: Finding a term when the terms are nonconsecutive**

If you know any two nonconsecutive terms of a geometric sequence, you can use this information to find the general formula of the sequence as well as any specified term. For example, if the 5th term of a geometric sequence is 64 and the 10th term is 2, you can find the 15th term. Just follow these steps:

**1.** **Determine the value of r.**

You can use the geometric formula to create a system of two formulas to find *r*: *a _{5}* =

*a*·

_{1}*r*

^{5 – 1}and

*a*=

_{10}*a*·

_{1}*r*

^{10 – 1}, or

You can use substitution to solve one equation for *a*_{1} (see Chapter 13 for more on this method of solving systems):

Plug this expression in for *a _{1}* in the other equation:

Now simplify this equation:

2 = 64*r*^{5}

2/64 = 1/32 = *r*^{5}

1/2 = *r*

**2.** **Find the specific formula for the given sequence.**

**a.** **Plug r into one of the equations to find a_{1}.**

This step gives you

**b.** **Plug a_{1} and r into the formula.**

Now that you know *a _{1}* and

*r,*you can write the formula:

*a _{n}* = 1,024(1/2)

^{(n – 1)}.

**3.** **Find the term you’re looking for.**

In this case, you want to find the 15th term (*n* = 15):

*a*_{15} = 1,024(1/2)^{15 – 1}

= 1,024(1/2)^{14}

= 1,024(1/16,384)

= 1/16

The annual depreciation of a car’s value is approximately 30 percent. Every year, the car is actually worth 70 percent of its value from the year before. If *a _{1}* represents the value of a car when it was new and

*n*represents the number of years that have passed,

*a*=

_{n}*a*· (0.7)

_{1}*when*

^{n}*n*≥ 0. Notice that this sequence starts at 0, which is okay as long as the information says that it starts at 0.

**Creating a Series: Summing Terms of a Sequence**

A *series* is the sum of terms in a sequence. Except for one situation where you can add the sum of an infinite series, you may be asked to find the sum of a certain number of terms (the first 12, for example). Summing a sequence is especially helpful in calculus when you begin discussing integration. Before some of the newer calculus concepts were discovered, mathematicians used series to find the areas under curves. Finding the area of a rectangle was easy, but because curves aren’t straight, finding the area under them wasn’t as easy. So they broke up the region into very small rectangles and added them together. This concept then evolved into an integral, and you will see a ton of that in calculus.

**Reviewing general summation notation**

The sum of the first *k* terms of a sequence is referred to as the* kth partial sum.* They’re called partial sums because you’re only able to find the sum of a certain number of terms — no infinite series here! You may use partial sums when you want to find the area under a curve (graph) between two certain values of *x.* Although finding the *entire* area under the graph isn’t usually possible (because it could be infinite if the curve goes on forever), you can find the area underneath a piece of it.

Don’t let the use of a different variable here confuse you. Instead of *k, *your book may still even use *n* and call it an *n*th partial sum. Remember that a variable just stands in for an unknown, so it really can be any variable you want — even those Greek variables that we used in the trig chapters. But we’ve most often seen books use *k* to represent the number of terms in a series and *n* for the number of terms in a sequence.

The notation of the *k*th partial sum of a sequence is as follows:

You read this equation as “the *k*th partial sum of *a _{n} *is . . .” where

*n*= 1 is the

*lower limit*of the sum and

*k*is the

*upper limit*of the sum. To find the

*k*th partial sum, you begin by plugging the lower limit into the general formula and continue in order, plugging in integers until you reach the upper limit of the sum. At that point, you simply add all the terms to find the sum.

To find the fifth partial sum of *a _{n} *=

*n*

^{3}– 4

*n*+ 2, for example, follow these steps:

**1. Plug all values of n (starting with 1 and ending with k) into the formula.**

Because you want to find the fifth partial sum, plug in 1, 2, 3, 4, and 5:

• *a _{1}* = (1)

^{3}– 4(1) + 2 = 1 – 4 + 2 = –1

• *a _{2}* = (2)

^{3}– 4(2) + 2 = 8 – 8 + 2 = 2

• *a _{3}* = (3)

^{3}– 4(3) + 2 = 27 – 12 + 2 = 17

• *a _{4}* = (4)

^{3}– 4(4) + 2 = 64 – 16 + 2 = 50

• *a _{5}* = (5)

^{3}– 4(5) + 2 = 125 – 20 + 2 = 107

**2.** **Add all the values from a_{1} to a_{k} to find the sum.**

This step gives you

–1 + 2 + 17 + 50 + 107 = 175

**3.** **Rewrite the final answer, using summation notation.**

**Summing an arithmetic sequence**

The *k*th partial sum of an arithmetic sequence still calls for you to add the first *k* terms. But in the arithmetic sequence, you do have a formula to use instead of plugging in each of the values for *n.* The *k*th partial sum of an arithmetic series is

You simply plug the lower and upper limits into the formula for *a _{n} *to find

*a*and

_{1}*a*

_{k}.One real-world application of an arithmetic sum involves stadium seating. Say, for example, a stadium has 35 rows of seats; there are 20 seats in the first row, 21 seats in the second row, 22 seats in the third row, and so on. How many seats do all 35 rows contain? Follow these steps to find out:

**1.** **Find the first term of the sequence.**

The first term of this sequence (or the number of seats in the first row) is given: 20.

**2.** **Find the kth term of the sequence.**

Because the stadium has 35 rows, find *a _{35}.* Use the formula for the

*n*th term of an arithmetic sequence (see the earlier section “Covering the Distance between Terms: Arithmetic Sequences”). The first term is 20, and each row has one more seat than the row before it, so

*d*= 1. Plug these values into the formula:

*a _{35}* =

*a*+ (35 – 1)

_{1}*d*

= 20 + (34) · 1

= 54

** Note:** This solution is the number of seats in the 35th row, not the answer to how many seats the stadium contains.

**3.** **Use the formula for the kth^{ }partial sum of an arithmetic sequence to find the sum.**

**Seeing how a geometric sequence adds up**

Just like when you found the sum of an arithmetic sequence, you can find the sum of a geometric sequence, which has the general explicit expression of *a _{n}*

_{ }=

*a*·

_{1}*r*

^{n}^{ – 1}. Also, because the formulas to find specific terms in the two types of sequences are different, so is the formula to find their sums.

Here we show you how to find the sum of two different types of geometric sequences. The first type is a finite sum (comparable to a *k*th partial sum from the previous section), and it too has an upper limit and a lower limit. The common ratio of partial sums of this type has no specific restrictions. The second type of geometric sum is called an *infinite* geometric sum, and the common ratio for this type is very specific (it *must* be strictly between –1 and 1). This type of geometric sequence is very helpful if you drop a ball and count how far it travels up and down, and then up and down, until it finally starts rolling.

Cars depreciate at an annual rate of 30 percent, starting the second you drive your new, shiny car off the lot. Say you originally pay $22,500 for a car; you can use the rate of depreciation and the price to figure out how much your car is worth at any given time — all by using geometric sequences. Just find the common ratio (which is the percent of the car that remains when the depreciation has been taken away) as a decimal. Using the original price as the first term, when *t* = 0 (because it’s brand new), you can use a geometric sequence to find out how much the car is worth after *t* years.

By definition, a geometric series continues infinitely, for as long as you want to keep plugging in values for *n.* However, in a specific type of geometric series, no matter how long you plug in values for *n,* the sum never gets larger than a certain value. This type of series has a specific formula to find the infinite sum. The sum isn’t infinite; the number of terms is. In mathematical terms, you say that some geometric sequences — ones with a common ratio between –1 and 1 — have a limit to their sequence of partial sums. In other words, the partial sum comes closer and closer to a particular number without ever actually reaching it. You call this number the *sum of thesequence,* as opposed to the *k*th partial sum you find in previous sections in this chapter.

**Stop right there: Determining the partial sum of a finite geometric sequence**

You can find a partial sum of a geometric sequence by using the following formula:

For example, follow the steps to find

**1.** **Find a_{1} by plugging in 1 for n.**

9(–1/3)^{1 – 1} = 9(1) = 9

**2.** **Find a_{2} by plugging in 2 for n.**

9(–1/3)^{2 – 1} = 9(–1/3)^{1} = –3.

**3.** **Divide a_{2} by a_{1} to find r.**

For this example, *r* = –3/9 = –1/3. Notice that this value is the same as the fraction in the parentheses.

You may have noticed that 9(–1/3)^{n}^{ – 1} follows the general formula for *a _{n}* =

*a*·

_{1}*r*

^{n}^{ – 1}(the general formula for a geometric sequence) exactly, where

*a*= 9 and

_{1}*r*= –1/3. However, if you didn’t notice it, the method used in Steps 1–3 works to a tee.

**4.** **Plug a_{1}, r, and k into the sum formula.**

The problem now boils down to the following simplifications:

•

•

Geometric summation problems take quite a bit of work with fractions, so make sure to find a common denominator, invert, and multiply when necessary. Or you can use a calculator and then reconvert to a fraction. Just be careful to use correct parentheses when entering the numbers.

**To geometry and beyond: Finding the value of an infinite sum**

Finding the value of an infinite sum in a geometric sequence is actually quite simple — as long as you keep your fractions and decimals straight. If *r* lies outside the range –1 < *r* < 1, *a _{n} *grows without bound infinitely, so there’s no limit on how large the absolute value of

*a*(|

_{n}*a*|) can get. If |

_{n}*r*| < 1, for every value of

*n,*|

*r*| continues to decrease infinitely until it becomes arbitrarily close to 0. This decrease is because when you multiply a fraction between –1 and 1 by itself, the absolute value of that fraction continues to get smaller until it becomes so small that you hardly notice it. Therefore, the term

^{n}*r*almost disappears completely in the finite geometric sum formula:

^{k}And if the *r ^{k} *disappears — or gets very small — the finite formula changes to the following and allows you to find the sum of an infinite geometric series:

For example, follow the steps to find this value:

**1.** **Find the value of a_{1} by plugging in 1 for n.**

*a _{1}* = 4(2/5)

^{1 – 1}

= 4(2/5)^{0}

= 4 · 1

= 4

**2.** **Calculate a_{2} by plugging in 2 for n.**

*a _{2}* = 4(2/5)

^{2 – 1}

= 4(2/5)^{1}

= 8/5

**3.** **Determine r.**

To find *r,* you divide *a _{2}* by

*a*

_{1}:.

**4.** **Plug a_{1} and r into the formula to find the infinite sum.**

Plug in and simplify to find the following:

Repeating decimals also can be expressed as infinite sums. Consider the number 0.5555555. . . . You can write this number as 0.5 + 0.05 + 0.005 + . . . , and so on forever. The first term of this sequence is 0.5; to find *r,*0.05 ÷ 0.5 = 0.1. Plug these values into the infinite sum formula:

This sum is finite only if *r* lies strictly between –1 and 1.

**Expanding with the Binomial Theorem**

A *binomial* is a polynomial with exactly two terms. Expressing the multiplication of binomials without any parentheses is called *binomial expansion*. Using the binomial theorem requires you to find the coefficients of this expansion.

Expanding many binomials takes a rather extensive application of the distributive property and quite a bit of time. Multiplying two binomials is easy if you use the FOIL method (see Chapter 4), and multiplying three binomials doesn’t take much more effort. Multiplying ten binomials, however, takes long enough that you may end up quitting short of the halfway point. And if you make a mistake somewhere along the line, it snowballs and affects every subsequent step.

Therefore, in the interest of saving bushels of time and energy, we present to you the binomial theorem. If you need to find the entire expansion for a binomial, this theorem is the greatest thing since sliced bread:

This formula gives you a very abstract view of how to multiply a binomial *n* times. It’s quite hard to read, actually. But this form is the way your textbook shows it to you.

We promise, the actual use of this formula is not as hard as it looks. Each

comes from a combination formula and gives you the coefficients for

each term (they’re sometimes called *binomial coefficients*). We tell you how

to deal with in the later section “Using algebra.”

For example, to find (2*y* – 1)^{4}, you start off the binomial theorem by replacing *a* with 2*y, b* with –1, and *n* with 4 to get:

You then have to simplify this mess. We break it down in the next sections. First we take a closer look at the binomial theorem; then we see how to find those dreaded binomial coefficients; and last (but certainly not least) we explore how to put all the parts together to get the final answer.

**Breaking down the binomial theorem**

The binomial theorem looks extremely intimidating, but it becomes much simpler if you break it down into smaller steps and examine the parts. Allow us to point out a few things to be aware of so that you don’t get confused somewhere along the way; after you have all this info straightened out, your task will seem much more manageable:

The binomial coefficients won’t necessarily be the coefficients in

your final answer. You’re raising each monomial to a power, including any coefficients attached to each of them.

The theorem is written as the sum of two monomials, so if your task is to expand the difference of two monomials, the terms in your final answer should alternate between positive and negative numbers.

The exponent of the first monomial begins at *n* and decreases by 1 with each sequential term until it reaches 0 at the last term. The exponent of the second monomial begins at 0 and increases by 1 each time until it reaches *n* at the last term.

The exponents of both monomials add to *n* — unless the monomials themselves have powers greater than 1.

**Starting at the beginning: Binomial coefficients**

Depending on how many times you must multiply the same binomial — a value also known as an *exponent* — the coefficients for that particular exponent are always be the same. The binomial coefficients are found by

using the combinations formula. If the exponent is relatively small,

you can use a shortcut called *Pascal’s triangle *to find these coefficients. If not, you can always rely on algebra!

**Using Pascal’s triangle**

*Pascal’s triangle,* named after the famous mathematician Blaise Pascal, names the coefficients for a binomial expansion. It is especially useful with lower degrees. For example, if a sadistic teacher asked you to find (3*x* + 4)^{10}, we wouldn’t recommend using this shortcut; instead, you’d just use the formula as described in the next section, “Using algebra.” Figure 14-1 illustrates this concept. Each row gives the coefficients to (*a* + *b*)* ^{n}*, starting with

*n*= 0, depending on the exponent. To find any row of the triangle, you always start with the beginning. The top number of the triangle is 1, as well as all the numbers on the outer sides. To get any term in the triangle, you find the sum of the two numbers above it.

**Figure 14-1:**Determining coefficients with Pascal’s triangle.

For instance, the binomial coefficients for (*a* + *b*)^{5} are 1, 5, 10, 10, 5, and 1 — in that order.

**Using algebra**

If you need to find the coefficients of binomials algebraically, we offer a formula for that as well. The *r*th coefficient for the *n*th binomial expansion is written in the following form:

You may recall the term *factorial* from your earlier math classes. If not, allow us to remind you: *n*!, read as “n factorial,” is defined as

1 · 2 · 3 · . . . · (*n* – 2) · (*n* – 1) · *n*

You read the expression for the binomial coefficient as “*n* choose *r.*”

You usually can find a button for combinations on a calculator. If not, you can use the factorial button and do each part separately.

To make things a little easier, 0! is defined as 1. Therefore, you have these equalities:

For example, to find the binomial coefficient given by , substitute the values into the formula:

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**Expanding by using the binomial theorem**

Using the binomial theorem can save you time, but it can be dangerous (the whole “nothing in life comes easy” philosophy). Keeping each of the steps separate until the very end should help. The final outcomes of a binomial expansion depend, as well, on whether the original monomial had no coefficients or exponents (other than 1) on the variables — we show you how to use the theorem in the next section. When the original monomial has coefficients or exponents other than 1 on the variable(s), you have to be careful to take those into account. We show you an example of that as well in the section “Raising monomials to a power pre-expansion.”

**Normal expansion problems**

To find the expansion of binomials with the theorem in a basic situation, follow these steps:

**1.** **Write out the binomial expansion by using the theorem, changing the variables where necessary.**

For example, consider the problem (*m* + 2)^{4}. According to the theorem, you should replace the letter *a* with *m,* the letter *b* with 2, and the exponent *n* with 4:

The exponents of *m* begin at 4 and end at 0 (see the section “Breaking down the binomial theorem”). Similarly, the exponents of 2 begin at 0 and end at 4. For each term, the sum of the exponents in the expansion is always 4.

**2.** **Find the binomial coefficients (see the section “Starting at the beginning: Binomial coefficients”).**

We use the combinations formula to find the five coefficients, but you could use Pascal’s triangle as a shortcut because the degree is so low (it wouldn’t hurt you to write out five rows of Pascal’s triangle — starting with 0 through 4).

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You may have noticed that after you reach the middle of the expansion, the coefficients are a mirror image of the first half. This trick is a time-saver you can employ so you don’t need to do all the calculations

for .

**3. Replace all with the coefficients from Step 2.**

This step gives you

1(*m*)^{4}(2)^{0} + 4(*m*)^{3}(2)^{1} + 6(*m*)^{2}(2)^{2} + 4(*m*)^{1}(2)^{3} + 1(*m*)^{0}(2)^{4}

**4.** **Raise the monomials to the powers specified for each term.**

1 · *m*^{4} · 1 + 4 · *m*^{3} · 2 + 6 · *m*^{2} · 4 + 4 · *m* · 8 + 1 · 1 · 16

**5.** **Combine like terms and simplify.**

*m*^{4} + 8*m*^{3} + 24*m*^{2} + 32*m* + 16

Notice that the coefficients you get in the final answer aren’t the binomial coefficients you find in Step 1. This difference is because you must raise each monomial to a power (Step 4), and the constant in the original binomial changed each term.

**Raising monomials to a power pre-expansion**

At times, monomials can have coefficients and/or be raised to a power before you begin the binomial expansion. In this case, you have to raise the entire monomial to the appropriate power in each step. For example, here’s how you expand the expression (3*x*^{2} – 2*y*)^{7}:

**1.** **Write out the binomial expansion by using the theorem, changing the variables where necessary.**

Replace the letter *a* in the theorem with the quantity (3*x*^{2}) and the letter *b* with (–2*y*). Don’t let those coefficients or exponents scare you — you’re still substituting them into the binomial theorem. Replace *n* with 7. You end up with

**2.** **Find the binomial coefficients (see the section “Starting at the beginning: Binomial coefficients”).**

Using the combination formula gives you the following:

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**3. Replace all with the coefficients from Step 2.**

1(3*x*^{2})^{7}(–2*y*)^{0} + 7(3*x*^{2})^{6}(–2*y*)^{1} + 21(3*x*^{2})^{5}(–2*y*)^{2} + 35(3*x*^{2})^{4}(–2*y*)^{3} + 35(3*x*^{2})^{3}(–2*y*)^{4} + 21(3*x*^{2})^{2}(–2*y*)^{5} + 7(3*x*^{2})^{1}(–2*y*)^{6} + 1(3*x*^{2})^{1}(–2*y*)^{7}

**4.** **Raise the monomials to the powers specified for each term.**

1(2,187*x*^{14})(1) + 7(729*x*^{12})(–2*y*) + 21(243*x*^{10})(4*y*^{2}) + 35(81*x*^{8})(–8y^{3}) + 35(27*x*^{6})(16*y*^{4}) + 21(9*x*^{4})(–32*y*^{5}) + 7(3*x*^{2})(64*y*^{6}) + 1(1)(–128*y*^{7})

**5.** **Simplify.**

2,187*x*^{14} – 10,206*x*^{12}*y* + 20,412*x*^{10}*y*^{2} – 22,680*x*^{8}*y*^{3} + 15,120*x*^{6}*y*^{4} – 6,048*x*^{4}*y*^{5} + 1,344*x*^{2}*y*^{6} – 128*y*^{7}

**Expansion with complex numbers**

The most complicated type of binomial expansion involves the complex number *i,* because you’re not only dealing with the binomial theorem but dealing with imaginary numbers as well. (For more on complex numbers, see Chapter 11.) When raising complex numbers to a power, note that *i*^{1}* = i,* *i*^{2} = –1, *i*^{3} = *–i,* and *i*^{4} = 1. If you run into higher powers, this pattern repeats: *i*^{5} = *i*, *i*^{6} = –1, *i*^{7} = –*i,* and so on. Because powers of the imaginary number *i* can be simplified, your final answer to the expansion should not include powers of *i.* Instead, use the information given here to simplify the powers of *i* and then combine your like terms.

For example, to expand (1 + 2*i*)^{8}, follow these steps:

**1.** **Write out the binomial expansion by using the theorem, changing the variables where necessary.**

(1 + 2*i*)^{8} expands to

**2.** **Find the binomial coefficients.**

Using the combination formula gives you the following:

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**3. Replace all with the coefficients from Step 2.**

1(1)^{8}(2*i*)^{0} + 8(1)^{7}(2*i*)^{1} + 28(1)^{6}(2*i*)^{2} + 56(1)^{5}(2*i*)^{3} + 70(1)^{4}(2*i*)^{4} + 56(1)^{3}(2*i*)^{5} + 28(1)^{2}(2*i*)^{6} + 8(1)^{1}(2*i*)^{7} + 1(1)^{0}(2*i*)^{8}

**4.** **Raise the monomials to the powers specified for each term.**

1(1)(1) + 8(1)(2*i*) + 28(1)(4*i*^{2}) + 56(1)(8*i*^{3}) + 70(1)(16*i*^{4}) + 56(1)(32*i*^{5}) + 28(1)(64*i*^{6}) + 8(1)(128*i*^{7}) + 1(1)(256*i*^{8})

**5.** **Simplify any i’s that you can.**

1(1)(1) + 8(1)(2*i*) + 28(1)(4)(–1) + 56(1)(8)(*–i*) + 70(1)(16)(1) + 56(1)(32)(*i*) + 28(1)(64)(–1) + 8(1)(128)(*–i*) + 1(1)(256)(1)

**6.** **Combine like terms and simplify.**

1 + 16*i* – 112 – 448*i* + 1,120 + 1,792*i* – 1,792 – 1,024*i* + 256

= –527 + 336*i*