Binomial Theorem

Learning Objectives

In this section, you will:

• Apply the Binomial Theorem.

A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find (x+y)n(𝑥+𝑦)𝑛 without multiplying the binomial by itself n𝑛 times.

Identifying Binomial Coefficients

In Counting Principles, we studied combinations. In the shortcut to finding (x+y)n,(𝑥+𝑦)𝑛, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation (nr)(𝑛𝑟) instead of C(n,r),𝐶(𝑛,𝑟), but it can be calculated in the same way. So

(nr)=C(n,r)=n!r!(n−r)!(𝑛𝑟)=𝐶(𝑛,𝑟)=𝑛!𝑟!(𝑛−𝑟)!

The combination (nr)(𝑛𝑟) is called a binomial coefficient. An example of a binomial coefficient is (52)=C(5,2)=10.(52)=𝐶(5,2)=10.

BINOMIAL COEFFICIENTS

If n𝑛 and r𝑟 are integers greater than or equal to 0 with n≥r,𝑛≥𝑟, then the binomial coefficient is

(nr)=C(n,r)=n!r!(n−r)!(𝑛𝑟)=𝐶(𝑛,𝑟)=𝑛!𝑟!(𝑛−𝑟)!

Q&A

Is a binomial coefficient always a whole number?

Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.

EXAMPLE 1

Finding Binomial Coefficients

Find each binomial coefficient.

1. ⓐ (53)(53)
2. ⓑ (92)(92)
3. ⓒ (97)(97)

Solution

Use the formula to calculate each binomial coefficient. You can also use the nCr𝑛𝐶𝑟 function on your calculator.

(nr)=C(n,r)=n!r!(n−r)!(𝑛𝑟)=𝐶(𝑛,𝑟)=𝑛!𝑟!(𝑛−𝑟)!

1. ⓐ (53)=5!3!(5−3)!=5⋅4⋅3!3!2!=10(53)=5!3!(5−3)!=5⋅4⋅3!3!2!=10
2. ⓑ (92)=9!2!(9−2)!=9⋅8⋅7!2!7!=36(92)=9!2!(9−2)!=9⋅8⋅7!2!7!=36
3. ⓒ (97)=9!7!(9−7)!=9⋅8⋅7!7!2!=36(97)=9!7!(9−7)!=9⋅8⋅7!7!2!=36

Analysis

Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.

(nr)=(nn−r)(𝑛𝑟)=(𝑛𝑛−𝑟)

TRY IT #1

Find each binomial coefficient.

1. ⓐ (73)(73)
2. ⓑ (114)(114)

Using the Binomial Theorem

When we expand (x+y)n(𝑥+𝑦)𝑛 by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand (x+y)52,(𝑥+𝑦)52, we might multiply (x+y)(𝑥+𝑦) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

(x+y)2=x2+2xy+y2(x+y)3=x3+3x2y+3xy2+y3(x+y)4=x4+4x3y+6x2y2+4xy3+y4(𝑥+𝑦)2=𝑥2+2𝑥𝑦+𝑦2(𝑥+𝑦)3=𝑥3+3𝑥2𝑦+3𝑥𝑦2+𝑦3(𝑥+𝑦)4=𝑥4+4𝑥3𝑦+6𝑥2𝑦2+4𝑥𝑦3+𝑦4

First, let’s examine the exponents. With each successive term, the exponent for x𝑥 decreases and the exponent for y𝑦 increases. The sum of the two exponents is n𝑛 for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

(n0),(n1),(n2),…,(nn).(𝑛0),(𝑛1),(𝑛2),…,(𝑛𝑛).

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

(x+y)n=∑k=0n(nk)xn−kyk=xn+(n1)xn−1y+(n2)xn−2y2+…+(nn−1)xyn−1+yn(𝑥+𝑦)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑥𝑛−𝑘𝑦𝑘=𝑥𝑛+(𝑛1)𝑥𝑛−1𝑦+(𝑛2)𝑥𝑛−2𝑦2+…+(𝑛𝑛−1)𝑥𝑦𝑛−1+𝑦𝑛

Another way to see the coefficients is to examine the expansion of a binomial in general form, x+y,𝑥+𝑦, to successive powers 1, 2, 3, and 4.

(x+y)1=x+y(x+y)2=x2+2xy+y2(x+y)3=x3+3x2y+3xy2+y3(x+y)4=x4+4x3y+6x2y2+4xy3+y4(𝑥+𝑦)1=𝑥+𝑦(𝑥+𝑦)2=𝑥2+2𝑥𝑦+𝑦2(𝑥+𝑦)3=𝑥3+3𝑥2𝑦+3𝑥𝑦2+𝑦3(𝑥+𝑦)4=𝑥4+4𝑥3𝑦+6𝑥2𝑦2+4𝑥𝑦3+𝑦4

Can you guess the next expansion for the binomial (x+y)5?(𝑥+𝑦)5?

Figure 1

See Figure 1, which illustrates the following:

• There are n+1𝑛+1 terms in the expansion of (x+y)n.(𝑥+𝑦)𝑛.
• The degree (or sum of the exponents) for each term is n.𝑛.
• The powers on x𝑥 begin with n𝑛 and decrease to 0.
• The powers on y𝑦 begin with 0 and increase to n.𝑛.
• The coefficients are symmetric.

To determine the expansion on (x+y)5,(𝑥+𝑦)5, we see n=5,𝑛=5, thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of x,𝑥, the pattern is as follows:

• Introduce x5,𝑥5, and then for each successive term reduce the exponent on x𝑥 by 1 until x0=1𝑥0=1 is reached.
• Introduce y0=1,𝑦0=1, and then increase the exponent on y𝑦 by 1 until y5𝑦5 is reached.x5,x4y,x3y2,x2y3,xy4,y5𝑥5,𝑥4𝑦,𝑥3𝑦2,𝑥2𝑦3,𝑥𝑦4,𝑦5

The next expansion would be

(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5.(𝑥+𝑦)5=𝑥5+5𝑥4𝑦+10𝑥3𝑦2+10𝑥2𝑦3+5𝑥𝑦4+𝑦5.

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal’s Triangle, shown in Figure 2. Pascal didn’t invent the triangle. The underlying principles had been developed and written about for over 1500 years, first by the Indian mathematician (and poet) Pingala in the second century BCE. Others throughout Asia and Europe worked with the concepts throughout, and the triangle was first published in its graphical form by Omar Khayyam, an Iranian mathematician and astronomer, for whom the triangle is named in Iran. French mathematician Blaise Pascal repopularized it when he republished it and used it to solve a number of probability problems.

Figure 2

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3^rd row, flank the ends of the rows with 1’s, and add 1+11+1 to find the middle number, 2. In the nth𝑛th row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

THE BINOMIAL THEOREM

The Binomial Theorem is a formula that can be used to expand any binomial.

(x+y)n=∑k=0n(nk)xn−kyk=xn+(n1)xn−1y+(n2)xn−2y2+…+(nn−1)xyn−1+yn(𝑥+𝑦)𝑛=∑𝑘=0𝑛(𝑛𝑘)𝑥𝑛−𝑘𝑦𝑘=𝑥𝑛+(𝑛1)𝑥𝑛−1𝑦+(𝑛2)𝑥𝑛−2𝑦2+…+(𝑛𝑛−1)𝑥𝑦𝑛−1+𝑦𝑛

HOW TO

Given a binomial, write it in expanded form.

1. Determine the value of n𝑛 according to the exponent.
2. Evaluate the k=0𝑘=0 through k=n𝑘=𝑛 using the Binomial Theorem formula.
3. Simplify.

EXAMPLE 2

Expanding a Binomial

Write in expanded form.

1. ⓐ (x+y)5(𝑥+𝑦)5
2. ⓑ (3x−y)4(3𝑥−𝑦)4

Solution

1. ⓐSubstitute n=5𝑛=5 into the formula. Evaluate the k=0𝑘=0 through k=5𝑘=5 terms. Simplify.(x+y)5(x+y)5=(50)x5y0+(51)x4y1+(52)x3y2+(53)x2y3+(54)x1y4+(55)x0y5=x5+5x4y+10x3y2+10x2y3+5xy4+y5(𝑥+𝑦)5=(50)𝑥5𝑦0+(51)𝑥4𝑦1+(52)𝑥3𝑦2+(53)𝑥2𝑦3+(54)𝑥1𝑦4+(55)𝑥0𝑦5(𝑥+𝑦)5=𝑥5+5𝑥4𝑦+10𝑥3𝑦2+10𝑥2𝑦3+5𝑥𝑦4+𝑦5
2. ⓑSubstitute n=4𝑛=4 into the formula. Evaluate the k=0𝑘=0 through k=4𝑘=4 terms. Notice that 3×3𝑥 is in the place that was occupied by x𝑥 and that –y–𝑦 is in the place that was occupied by y.𝑦. So we substitute them. Simplify.(3x−y)4(3x−y)4=(40)(3x)4(−y)0+(41)(3x)3(−y)1+(42)(3x)2(−y)2+(43)(3x)1(−y)3+(44)(3x)0(−y)4=81×4−108x3y+54x2y2−12xy3+y4(3𝑥−𝑦)4=(40)(3𝑥)4(−𝑦)0+(41)(3𝑥)3(−𝑦)1+(42)(3𝑥)2(−𝑦)2+(43)(3𝑥)1(−𝑦)3+(44)(3𝑥)0(−𝑦)4(3𝑥−𝑦)4=81𝑥4−108𝑥3𝑦+54𝑥2𝑦2−12𝑥𝑦3+𝑦4

Analysis

Notice the alternating signs in part b. This happens because (−y)(−𝑦) raised to odd powers is negative, but (−y)(−𝑦) raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.

TRY IT #2

Write in expanded form.

1. ⓐ (x−y)5(𝑥−𝑦)5
2. ⓑ (2x+5y)3(2𝑥+5𝑦)3

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as (x+2y)16(𝑥+2𝑦)16 can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of (x+y)5.(𝑥+𝑦)5.

(x+y)5=x5+(51)x4y+(52)x3y2+(53)x2y3+(54)xy4+y5(𝑥+𝑦)5=𝑥5+(51)𝑥4𝑦+(52)𝑥3𝑦2+(53)𝑥2𝑦3+(54)𝑥𝑦4+𝑦5

The second term is (51)x4y.(51)𝑥4𝑦. The third term is (52)x3y2.(52)𝑥3𝑦2. We can generalize this result.

(nr)xn−ryr(𝑛𝑟)𝑥𝑛−𝑟𝑦𝑟

THE (R+1)TH TERM OF A BINOMIAL EXPANSION

The (r+1)th(𝑟+1)th term of the binomial expansion of (x+y)n(𝑥+𝑦)𝑛 is:

(nr)xn−ryr(𝑛𝑟)𝑥𝑛−𝑟𝑦𝑟

HOW TO

Given a binomial, write a specific term without fully expanding.

1. Determine the value of n𝑛 according to the exponent.
2. Determine (r+1).(𝑟+1).
3. Determine r.𝑟.
4. Replace r𝑟 in the formula for the (r+1)th(𝑟+1)th term of the binomial expansion.

EXAMPLE 3

Writing a Given Term of a Binomial Expansion

Find the tenth term of (x+2y)16(𝑥+2𝑦)16 without fully expanding the binomial.

Solution

Because we are looking for the tenth term, r+1=10,𝑟+1=10, we will use r=9𝑟=9 in our calculations.

(nr)xn−ryr(𝑛𝑟)𝑥𝑛−𝑟𝑦𝑟

(169)x16−9(2y)9=5,857,280x7y9(169)𝑥16−9(2𝑦)9=5,857,280𝑥7𝑦9

TRY IT #3

Find the sixth term of (3x−y)9(3𝑥−𝑦)9 without fully expanding the binomial.