Lesson 10-3 The Binomial Theorems3.amazonaws.com/scschoolfiles/695/631_smp_sefst_c10l03.pdf · 2016-02-02 · The Binomial Theorem 631 BIG IDEA The Binomial Theorem shows that the

The Binomial Theorem 631

BIG IDEA The Binomial Theorem shows that the coeffi cients of terms in the expansion (a + b)n are combinations that can be read from Pascal’s Triangle.

Recall that to expand a product or power of polynomials means to write the expression as a sum.

Step 1 Use the expand command on a CAS to expand (x + y)n for n = 1, 2, 3, 4, and 5.

Step 2 Write the coeffi cients as a table, with the coeffi cients of (x + y)1 in the fi rst row, of (x + y)2 in the second row, and so on, with the coeffi cients written in order of the decreasing powers of x. What patterns do you notice? Describe them as clearly and specifi cally as you can.

Step 3 Based on your answer to Step 2, predict what coeffi cients will result for (x + y)6. Use a CAS to check your prediction.

Step 4 Does (x + y)0 fi t into this pattern? Explain.

The Activity illustrates that the coeffi cients of the terms in the expansion of (x + y)n are the numbers in the nth row of Pascal’s Triangle. Consider (x + y)3 = (x + y)(x + y)(x + y). You can fi nd this product by multiplying each term of the fi rst factor by each term of each of the other factors and then adding those partial products. In all, eight different partial products are computed. The terms that are multiplied to get each partial product are highlighted in red at the right.Adding like terms gives (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3. The coeffi cients are the entries row 3 of Pascal’s Triangle.

ActivityActivity

Degree n Coeffi cients (decreasing powers of x)

1 1 12 1 2 ?3 ? ? ? ?4 ? ? ? ? ?5 ? ? ? ? ? ?6 ? ? ? ? ? ? ?

Degree n Coeffi cients (decreasing powers of x)

1 1 12 1 2 ?3 ? ? ? ?4 ? ? ? ? ?5 ? ? ? ? ? ?6 ? ? ? ? ? ? ?

Choice of Factors(in red) Partial Product

(x + y)(x + y)(x + y) xxx = x3

(x + y)(x + y)(x + y) xxy = x2y(x + y)(x + y)(x + y) xyx = x2y(x + y)(x + y)(x + y) xyy = xy2

(x + y)(x + y)(x + y) yxx = x2y(x + y)(x + y)(x + y) yxy = xy2

(x + y)(x + y)(x + y) yyx = xy2

(x + y)(x + y)(x + y) yyy = y3

Choice of Factors(in red) Partial Product

(x + y)(x + y)(x + y) xxx = x3

(x + y)(x + y)(x + y) xxy = x2y(x + y)(x + y)(x + y) xyx = x2y(x + y)(x + y)(x + y) xyy = xy2

(x + y)(x + y)(x + y) yxx = x2y(x + y)(x + y)(x + y) yxy = xy2

(x + y)(x + y)(x + y) yyx = xy2

(x + y)(x + y)(x + y) yyy = y3

Mental Math

Multiply.

a. (4 + 8)(4 + 8)

b. (x + y)(x + y)

c. (40 + 1)(40 + 1)

Vocabulary

expansion of (x+y)n

binomial coeffi cients

Lesson

The Binomial Theorem10-3

Lesson 10-3

SMP_SEFST_C10_L03_631_636_FINAL.631 631SMP_SEFST_C10_L03_631_636_FINAL.631 631 6/9/09 3:17:15 PM6/9/09 3:17:15 PM

632 Binomial Distributions

Chapter 10

To see why each coeffi cient is a combination, examine the eight partial products. Each partial product is a sequence of x’s and y ’s resulting from choosing x’s from some factors and y ’s from other factors. The result x3 occurs when x is used as a factor three times and y is used as a factor zero times. There are 3C0 ways to choose 0 y ’s from the three y terms, so x3 occurs in 3C0 = 1 way. That is, x3 occurs once, so its coeffi cient in the expansion is 1.

The product x2y occurs when x is chosen from two of the three factors and y from one. There are three y terms from which to choose, so this can be done in 3C1 = 3 ways. So x2y occurs three times and its coeffi cient in the expansion is 3. Similarly, xy2 occurs when x is chosen from one factor and y from two, so the coeffi cient of xy2 is 3C2 = 3.

Finally, y3 occurs when all three y terms are chosen, and this occurs in

3C3 = 1 way. Thus,

(x + y) 3 = x3 + 3x2y + 3xy2 + y3.Using the language of combinations, we can write

(x + y) 3 = 3C0 x3 + 3C1x

2y + 3C2 xy2 + 3C3 y3.

Notice that each coeffi cient nCr can be viewed as the answer to a problem in counting strings. For instance, 3C1 is the number of ways 2 x’s and 1 y can be arranged.

What Is the Binomial Theorem?

In general, the expansion of (x + y)n has nC0 xn = xn as its fi rst term

and nCn yn = yn as its last. The second term is nC1x

n–1y = nxn–1y, and the second from the last is nCn–1xyn–1 = nxyn–1. The sum ofthe exponents in each term is n, and the coeffi cient of xn–kyk is nCk. As the preceding discussion suggests, an easy way to rememberthe coeffi cient is that nCk is the number of strings of x’s and y ’s in which y occurs exactly k times and x occurs exactly (n – k) times. These results are found in the following important theorem, fi rst proved by Omar Khayyam.

Binomial Theorem

For any nonnegative integer n, (x + y)n = nC0xny0 + nC1xn–1y1 + nC2xn–2y2 + ... +nCkx

n–kyk + ... + nCnx0yn

= ∑ k = 0

n

nCkxn–kyk .

For example, (x + y)4 = 4C0 x4 + 4C1x

3y + 4C2 x2y2 + 4C3 xy3 + 4C4 y

4

= x4 + 4x3y + 6x2y2 + 4xy3 + y4.

Because of their application in this theorem, the combinations nCk are sometimes called binomial coefficients.

Tomb of Omar Khayyam

(1048-1123) in Neishabur, Iran

Tomb of Omar Khayyam

(1048-1123) in Neishabur, Iran



Lesson 10-3

Some Uses of the Binomial Theorem

The Binomial Theorem can be used to expand the power of any binomial. Even if you have a CAS, it can sometimes be faster to use the Binomial Theorem to fi nd the coeffi cient of a specifi c term rather than to work out the entire expansion.

Example 1Find the power of y and the coeffi cient of the x3 term in (x + y)8.

Solution From the eight factors of (x + y)8, x is to be chosen three times, so y will be chosen ? times, and the power of y is ? . Three x’s can be chosen in 8C ?

= ? ways, so the coeffi cient of the x3 term is ? .

Check Use a CAS to expand (x + y)8.

QY1

Recall from Lesson 7-7 that chunking is the process of grouping small bits of information into a single piece of information. That method is useful when applying the Binomial Theorem.

Example 2Expand (3w – 2)4 without a CAS.

Solution Use the Binomial Theorem or Pascal’s Triangle to expand (x + y)4.

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

Let x = 3w and y = –2 and substitute.

(3w - 2)4 = (3w)4 + 4(3w)3(–2) + 6(3w)2(–2)2 + 4(3w)(–2)3 + (–2)4

= 81w4 + 4(27)(–2)w3 + 6(9)(4)w2 + 4(3)(–8)w + 16

= 81w4 - 216w3 + 216w2 - 96w + 16

Check Let w = 1. Then (3w – 2)4 = 14 = 1. The right side of the equation is 81 – 216 + 216 – 96 + 16 = 1. This checks the coeffi cients.

QY2

As you have seen, the Binomial Theorem links counting problems to algebra.

GUIDEDGUIDED

QY1

Find the coeffi cient of x4y2 in (x + y)6.

QY1

Find the coeffi cient of x4y2 in (x + y)6.

QY2

Find the coeffi cient of x3y3 in (x – 2y)6.

QY2

Find the coeffi cient of x3y3 in (x – 2y)6.



Chapter 10

Example 3a. A coin is fl ipped fi ve times. How many of the possible arrangements of

heads and tails have at least two heads?

b. Find all terms in (H + T)5 in which the power of H is at least two.

Solution

a. “At least two” means there could be 2, 3, 4, or 5 heads. Count according to the number of heads. Make a table to organize your work.

b. Calculate coeffi cients according to the power of H.

The terms in which the power of H is at least two are 10H2T3, 10H3T2, 5H4T, and H5.

Check Of the total 25 = 32 possible arrangements, 1 has no heads and 5 have one head, so 6 have fewer than 2 heads. That leaves 32 - 6 = 26 outcomes with at least 2 heads. It checks.

Example 3 shows that the coeffi cient of H rT

n–r in the expansion of (H + T )n is the number of ways of obtaining exactly r heads in n fl ips of a coin.

A pretty penny The U.S.

Mint estimates that a coin

stays in circulation for at

least 30 years.

A pretty penny The U.S.

Mint estimates that a coin

stays in circulation for at

least 30 years.

Number of Heads

Number of Tails

Sequence Type Arrangements

2 3 HHTTT 5C2 = 10

3 2 HHHTT 5C3 = 10

4 1 HHHHT 5C4 = 5

5 0 HHHHH 5C5 = 1

Total 26

Number of Heads

Number of Tails

Sequence Type Arrangements

2 3 HHTTT 5C2 = 10

3 2 HHHTT 5C3 = 10

4 1 HHHHT 5C4 = 5

5 0 HHHHH 5C5 = 1

Total 26

Power of H Power of T Product Type Coeffi cients

2 3 H2T35C2 = 10

3 2 H3T25C3 = 10

4 1 H4T 5C4 = 5

5 0 H55C5 = 1

Power of H Power of T Product Type Coeffi cients

2 3 H2T35C2 = 10

3 2 H3T25C3 = 10

4 1 H4T 5C4 = 5

5 0 H55C5 = 1



Lesson 10-3

Questions

COVERING THE IDEAS

1. a. In the product (x + y)(x + y)(x + y)(x + y), what is the coeffi cient of x3y?

b. How can this coeffi cient be derived from a combination problem?

2. Use the Binomial Theorem to fi nd the coeffi cient of x4 in (x – 3)7.

3. How many ways are there to get at least 4 heads in 7 fl ips of a coin?

4. a. Write an expression for the coeffi cient of x997 in (x + 1)1008. b. Find its value.

5. Find the term containing p3 in (6p + 2q)4.

In 6–8, expand with a CAS. Then check by letting x = 2 and evaluating

the given binomial power and the expansion.

6. (x – 2)6 7. (1 – 3x)4 8. (2x + 1)5

9. How many of the possible arrangements of heads and tails in six tosses of a coin have the following?

a. exactly 3 tails b. at least 3 tails

10. True or False Suppose kxpyq is a term in the expansion of (x + y)n. a. p + q = n b. k = pCq

APPLYING THE MATHEMATICS

11. Calculate 10014 by expanding (103 + 1)4 using the Binomial Theorem.

12. Calculate 0.94 by expanding (1 – 0.1)4 using the Binomial Theorem.

13. A coin is fl ipped six times. a. In how many ways is it possible to get exactly four heads? b. One such outcome is HHHHTT. If heads and tails are equally

likely on each fl ip, what is the probability of this outcome?

c. If the probability of heads is 3 _ 4 on each fl ip, what is the probability of the outcome HHHHTT?

14. Mr. and Mrs. Ippy hope to have four children. a. In how many orders can they have two boys and

two girls? b. Assuming boys and girls are equally likely, what is the

probability that they will have two boys and two girls? 3

_ 8



Chapter 10

15. Use the Binomial Theorem to prove Property 4 from Lesson 10-2,

that ∑ k = 0

n

nCk = 2n. Hint: Let 2n = (1 + 1)n.

16. a. Use a CAS to fi nd the coeffi cient of x3y2z2 in (x + y + z)7. b. What counting problem involving strings does this

coeffi cient answer?

REVIEW

In 17 and 18, use the numbers below, which are the fi rst six terms of

row 11 of Pascal’s Triangle. (Lesson 10-2)

1 11 55 165 330 462

17. Which term represents 11C5?

18. a. Write the rest of row 11. b. Write row 12 in full. c. Write row 10 in full.

In 19 and 20, suppose a charity sells 1000 raffl e tickets for $100 each.

a. Tell whether the question asks for a combination or

a permutation.

b. Answer the question. (Lessons 10-1, 6-8)

19. The fi rst place winner receives $25,000, the second place winner receives $15,000, the third place winner receives $10,000, and the fourth place winner receives $5,000. In how many different ways could the prizes be distributed?

20. All four winners share $55,000 evenly. How many ways could the prizes be distributed?

21. Three different integers are chosen at random from the integers 1 to 36. What is the probability that they are consecutive integers? (Lessons 10-1, 6-8)

EXPLORATION

22. a. Expand (1 + 0.001)5 to obtain the decimal for 1.0015. b. How many terms of the expansion are needed to get an estimate

to 1.0015 accurate to the nearest thousandth? c. Use your results from Parts a and b to estimate 1.0028 to the

nearest millionth. d. Give the complete decimal expansion of 1.0028. e. How close is your calculator value of 1.0028 to the value of your

answer in Part c?

QY ANSWERS

1. 15

2. –160

QY ANSWERS

1. 15

2. –160


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Lesson 10-3 The Binomial Theorems3.amazonaws.com/scschoolfiles/695/631_smp_sefst_c10l03.pdf · 2016-02-02 · The Binomial Theorem 631 BIG IDEA The Binomial Theorem shows that the