The thing that started it 8.6 THE BINOMIAL THEOREM

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476 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

(b) Based on your results for(a), guess the minimum number of moves required if you start with an arbi-trary number ofndisks. (Hint:To help see a pat-tern, add 1 to the number of moves forn51, 2, 3, 4.)

(c) A legend claims that monks in a remote monastery are working to move a set of 64 disks, and that the world will end when they complete their sacred task. Moving one disk per second without error, 24 hours a day, 365 days a year, how long would it take to move all 64 disks?

55. Number of Handshakes Suppose there arenpeople at a party and that each person shakes hands with every other person exactly once. Let f~n! denote the total

8.6 T H E B I N O M I A L T H E O R E M

We remake nature by the act of discovery, in the poem or in the theorem. And the great poem and the great theorem are new to every reader, and yet are his own experiences, because he himself recreates them.@And#in the instant when the mind seizes this for itself, in art or in science, the heart misses a beat.

J. Bronowski In this section we derive a general formula to calculate an expansion for~a1b!n

for any positive integer powern, or to find any particular term in such an expansion. We begin by calculating the first few powers directly and then look for significant patterns. To go from one power of ~a1b! to the next, we simply multiply by

~a1b!: ~a1b!15_a1 _b ~a1b!25_a21_2ab1_b2 ~a1b!35_a31_3a2_b1 _3ab21_b3 ~3! a1b a41_3a3_b1_3a2_b21_ab3

a3_b1_3a2_b21_3ab31 _b4 _{Add like terms}

~a1b!45_a41_4a3_b1 _6a2_b21_4ab31_b4 ~3! a1b a51 4a4 b16a3 b21 4a2 b31 ab4 a4 b14a3 b21 6a2 b31 4ab41

b5 _{Add like terms}

~a1b!55_a51_5a4_b1 _10a3_b21_10a2_b31_5ab41 _b5

T

he thing that started it all was this silly newspaper puzzle that asked you to count up the total number of ways you could spell the words “Pyramid of Values” from a triangular array of letters. This led my friend and me to discover Pascal’s triangle. This happened in grade 10 or 11.

Bill Gosper

number of handshakes. See Exercise 19, page 469. Show

f~n!5n~n21!

2 for every positive integern.

56. Supposenis an odd positive integer not divisible by 3. Show thatn2_{21 is divisible by 24. (Hint:Consider the} three consecutive integers n21, n, n11. Explain why the product~n21!~n11!must be divisible by 3 and by 8.)

57. Finding Patterns If an5Ï24n11, (a)write out the first five terms of the sequence$an%.(b)What odd integers occur in$an%?(c)Explain why$an%contains all primes greater than 3. (Hint:Use Exercise 56.)

When we look at these expansions of~a1 b!n_forn5_{1, 2, 3, 4, and 5, several}

patterns become apparent.

1. There aren1 1 terms, froman_tobn_.

2. Every term has essentially the same form: some coefficient times the product of a power ofatimes a power ofb.

3. In each term the sum of the exponents onaandbis alwaysn.

4. The powers (exponents) onadecrease, term by term, fromndown to 0 where the last term is given bybn₅

a0

bn

, and the exponents onb in-crease from 0 ton.

Knowing the form of the terms in the expansion and that the sum of the powers is alwaysn, we will have the entire expansion when we know how to calculate the coefficients of the terms. If we display the coefficients from the computations above, we find precisely the numbers in the first few rows of Pascal’s triangle:

1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

Using the address notation for Pascal’s triangle that we introduced in Section 8.4, the last row of coefficients in the triangle is~0

5! ,~1 5! ,~2 5! ,~3 5! ,~4 5! ,~5 5! , and so we may write the expansion for~a1b!5

: ~a1 b!55

S

5 0

D

a 5_b01

S

5 1

D

a 4_b11

S

5 2

D

a 3_b21

S

5 3

D

a 2_b3 1

S

5 4

D

a 1 b4₁

S

5 5

D

a 0 b5

Each term exhibits the same form. Forn5 5, each coefficient has the form~r5!,

whereris also the exponent onb.For each term the sum of the exponents onaand bis always 5, so that when we havebr_{, we must also havea}52r_{. Finally, since the}

first term hasr 50, the second term hasr 5 1, etc., the~r 11!st term involvesr. This leads to a general conjecture for the expansion of~a1 b!n

which we state as a theorem that can be proved using mathematical induction. (See the end of this section.)

Binomial theorem

Supposenis any positive integer. The expansion of~a1b!n_{is given by}

~a 1b!n5

S

n 0

D

a n_b01

S

n 1

D

a n21_b11_{· · ·}1

S

n r

D

a n2r_br1_{· · ·}1

S

n n

D

a 0_bn (1) where the~r 11!st term is

S

n

r

D

a n2r br , 0#r #n. In summation notation, ~a1 b!n5

o

n r50

S

n r

D

a n2r_br_{. (2)}

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478 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

At this point, we have established only that the form of our conjecture is valid for the first five values ofn, and we have not completely justified our use of the name Pascal’s triangle of binomial coefficients. Nonetheless, the multiplication of

~a1b!4_by~_a1 _b!_{to get the expansion for}~_a 1_b!5_{contains all the essential ideas}

of the proof.

We still lack a closed-form formula for the binomial coefficients. We know, for example, that the fourth term of the expansion of~x 12y!20

is~3 20_!

x17_~

2y!3

, but we cannot complete the calculation without the binomial coefficient~3

20_!

. This would require writing at least the first few terms of 20 rows of Pascal’s triangle.

Pascal himself posed and solved the problem of computing the entry at any given address within the triangle. He observed that to find~r

n!

, we can take the product of all the numbers from 1 throughr, and divide it into the product of the same number of integers, fromndownward. This leads to the following formula.

Pascal’s formula for binomial coefficients

Supposenis a positive integer andris an integer that satisfies 0,r #n. The binomial coefficient

S

n

r

D

is given by

S

n r

D

5

n~n21!· · ·~n 2r 11!

1 · 2 · 3 · · ·r (3)

We leave it to the reader to verify that the last factor in the numerator,

~n2r 11!, is therth number counting down fromn.This gives the same number of factors in the numerator as in the denominator.

cEXAMPLE 1 Using Pascal’s formula Find the first five binomial coefficients on the tenth row of Pascal’s triangle, and then give the first five terms of the expansion of~a1 b!10_.

Solution

Follow the strategy.

S

10 1

D

5 10 1 5 10,

S

10 2

D

5 10 · 9 1 · 2 545,

S

10 3

D

5 10 · 9 · 8 1 · 2 · 3 5120, and

S

10 4

D

5 10 · 9 · 8 · 7 1 · 2 · 3 · 4 5210.

Therefore the first five terms in the expansions of~a 1b!10_are

a10 1_10a9_b1_45a8_b21_120a7_b31 _210a6_b4_{. b}

There is another very common formula for binomial coefficients that uses factori-als. Equation (3) has a factorial in the denominator, and we can get a factorial in the numerator if we multiply numerator and denominator by the product of the rest of the integers from n2rdown to 1:

S

n r

D

5 n~n2 1!. . . ~n2 r 11! 1 · 2 · . . . r 5n~n2 1!. . . ~n2 r 11! r! · ~n2r!. . . 2 · 1 ~n2r!. . . 2 · 15 n! r! ~n2r!!. Strategy: We know~10 0!5 1. Use Equation (3) to get the remaining coefficients.

HISTORICAL NOTE

BLAISE PASCAL

While Equation (3) does not give a formula for (0n), the formulation in terms

of factorials does apply.

S

n 0

D

5 n! 0! ~n20!!5 n! 1 ·n! 5 n! n!51. This gives an alternative to Pascal’s formula.

Alternative formula for binomial coefficients

Supposenis a positive integer and ran integer that satisfies 0#r #n. The binomial coefficient~r n_! is given by

S

n r

D

5 n! r! ~n2 r!! (4)

Pascal’s triangle is named after Blaise Pascal, born in France in 1623. Pascal was an individual of incredible talent and breadth who made basic contributions in many areas of mathematics, but who died early after spending much of life embroiled in bitter philosophical and religious wrangling.

For some reason, Pascal’s father decided that his son should not be exposed to any mathematics. All mathematics books in the home were locked up and the subject was banned from discussion. We do not know if the appeal of the forbidden

was at work, but young Pascal

approached his father directly and asked what geometry was. His father’s answer so fascinated the 12-year-old boy that be began exploring geometric relationships on his own. He apparently rediscovered much of Euclid

completely on his own. When Pascal was introduced to conic sections ~see Chapter 10!he quickly absorbed everything available; he submitted a paper on conic sections to the French academy when he was only 16 years of age.

At the age of 29, Pascal had a conversion experience that led to a vow to renounced mathematics for a life of religious contemplation. Before that time, however, in addition to his foundational work in geometry, he built a mechanical computing machine (in honor of which the structured computer language Pascal is named), explored relations among binomial coefficients so thoroughly that we call the array of binomial

coefficients Pascal’s triangle even though the array had been known, at least in part, several hundred years earlier, proved the binomial theorem, gave the first published proof by mathematical induction, and invented (with Fermat) the science of combinatorial analysis, probability, and mathematical statistics.

Before his death ten years later, Pascal spent only a few days on mathematics. During a night made sleepless by a toothache, he concentrated on some problems about the cycloid curve that had attracted many mathematicians of the period. The pain subsided, and, in gratitude, Pascal wrote up his work for posterity.

Blaise Pascalmade

significant contributions to the study of mathematics before deciding to devote

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480 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers

cEXAMPLE 2 Symmetry in binomial coefficients Show that (a)

S

6 2

D

5

S

6 4

D

(b)

S

n r

D

5

S

n n2 r

D

. Solution

Follow the strategy.

Strategy: Use Equation (4)

to evaluate both sides of the _(a)

S

6

2

D

5 6! 2! ~622!! 5 6! 2! 4! and

S

6 4

D

5 6! 4! ~624!! 5 6! 4! 2!.

given equations to show that the two sides of each

equa-(b)

S

n r

D

5

n!

r! ~n2r!! and

tion are equal.

S

n n2r

D

5 n! ~n2 r!! @n2 ~n2r!#!5 n! ~n2r!!r! Thus

S

n r

D

5

S

n n 2r

D

. b

cEXAMPLE 3 Adding binomial coefficients Show that~3 8_{! 1 ~} 4 8_{!5 ~} 4 9_! . Get a common denominator and add fractions, but do not evaluate any of the factorials or binomial coefficients.

Solution

Use Equation (3) to get~38!and~48!, get common denominators, then add.

S

8 3

D

1

S

8 4

D

5 8 · 7 · 6 1 · 2 · 31 8 · 7 · 6 · 5 1 · 2 · 3 · 45 ~8 · 7 · 6!· 4 ~1 · 2 · 3!· 41 8 · 7 · 6 · 5 1 · 2 · 3 · 4 58 · 7 · 6~415! 1 · 2 · 3 · 4 5 9 · 8 · 7 · 6 1 · 2 · 3 · 45

S

9 4

D

. Thus

S

8 3

D

1

S

8 4

D

5

S

9 4

D

. b

Example 3 focuses more on the process than the particular result, hence the instruction to add fractions without evaluating. When we write out the binomial coefficients as fractions, we can identify the extra factors we need to get a common denominator and then add. In Example 2, we proved that (r

n₎ 5 ~ n2r

n ! _{giving a}

symmetry property for thenth row of Pascal’s triangle. Example 3 illustrates the essential steps to prove the following additivity property (see Exercise 61):

S

n r

D

1

S

n r 11

D

5

S

n11 r 11

D

Symmetry and additivity properties

Binomial coefficients have the following properties: Symmetry

S

n r

D

5

S

n n2r

D

(5) Additivity

S

n r

D

1

S

n r 11

D

5

S

n1 1 r 1 1

D

(6)

Notice that we used the additivity property from Equation (6) in Section 8.4 to get the ~n11!st row from thenth row in Pascal’s triangle. This justifies our claim that the entries in Pascal’s triangle are binomial coefficients.

Evaluating binomial coefficients TECHNOLOGY TIP

r

Most graphing calculators have the capacity to evaluate binomial coefficients directly, but we need to know where to look for the needed key. Most calculators use the notationnCr(meaning “number of combinations takenrat a time,” language from probability), and the key is located in a probability (PRB,PROB) submenu under theMATHmenu. To evaluate, say~20₄!, the process is as follows.

All TI-calculators: Having entered20on your screen, pressMATH PRB nCr, which putsnCr on the screen. Then type4so you have20nCr4. When you enter, the display should read4845.

All Casio calculators: Having entered20on your screen, pressMATH PRB nCr, which putsCon the screen. Then type4so you have 20C4. When you execute, the display should read4845.

HP-38: PressMATH, go down toPROB, highlightCOMB,OK. Then, on the command line you wantCOMB(20,4). Enter to evaluate.

HP-48: Put 20and4on the stack. As with many HP-48 operations,COMB (for “combinations”) works with two numbers.MTH NXT PROB COMBreturns4845.

cEXAMPLE 4 Binomial Theorem Use the binomial theorem to write out

the first five terms of the binomial expansion of~x 12y2!20_{and simplify.}

Solution

Use Equation (1) with a 5 x, b52y2

, and n520. The first five terms of

~x 12y2!20 are x201

S

20 1

D

x 19~_2y2!1

S

20 2

D

x 18~_2y2!21

S

20 3

D

x 17~_2y2!31

S

20 4

D

x 16~_2y2!4_.

Before simplifying, find the binomial coefficients, using either Equation (3) or the Technology Tip.

S

20 1

D

5 20 1 520

S

20 2

D

5 20 · 19 1 · 2 5190

S

20 3

D

5 20 · 19 · 18 1 · 2 · 3 5 1140

S

20 4

D

5 20 · 19 · 18 · 17 1 · 2 · 3 · 4 54845 Therefore, the first five terms of~x 12y2!20_are

x201_{20 · 2x}19_y2 1_{190 · 4x}18_y41_{1140 · 8x}17_y61_{4845 · 16x}16_y8_{, or} x201 40x19 y21 760x18 y41 9120x17 y61 77520x16 y8 . b

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482 Chapter 8 Discrete Mathematics Functions on the Set of Natural Numbers

cEXAMPLE 5 Finding a middle term In the expansion of~2x22 1

x! 10_{, find}

the middle term. Solution

There are 10 11 or 11 terms in the expansion of a tenth power, so the middle term is the sixth (five before and five after). The sixth term is given byr 55.

S

10 5

D

~2x 2!5

S

21 x

D

5 5252~32x10!

S

21 x

D

5 5 28064x5

The middle term is 28064x5

. b

cEXAMPLE 6 Finding a specified term In the expansion of (2x2₂

x 1₎10_{, find} the term whose simplified form involves 1

x.

Solution

Follow the strategy. The general term given in Equation (2) is

Strategy: First find the general term, then simplify.

Finally, find the value ofr

S

10

r

D

~2x 2!102r

S

21 x

D

r 5

S

10 r

D

2 102r_x2022r~2₁!r_x2r that gives21 as the

expo-nent ofx. 5

S

10 r

D

~21! r 2102r x2023r . For the term that involves 1

xorx

21_{, find the value ofrfor which the exponent}

onxis21: 20 23r 5 21, orr 57. The desired term is given by

S

10 7

D

~2x 2!3

S

21 x

D

7 5 2120~8!x6 x7 5 2 960 x . b

Proof of the Binomial Theorem

We can use mathematical induction to prove that Equation (1) holds for every positive integern.

(a) Forn5 1, Equation (1) is~a1b!15 ~ 0

1!_a1_b01 ~ 1

1!_a0_b15 _a1_{b, so}

Equa-tion (1) is valid whennis 1. (b) Hypothesis: ~a1 b!k 5

S

k 0

D

a k 1

S

k 1

D

a k21_b1_{· · ·} 1

S

k r

D

a k2r br₁ · · ·1

S

k k

D

b k (7) Conclusion: ~a 1b!k11₅

S

k 11 0

D

a k11₁

S

k1 1 1

D

a k_{b1· · · (8)} 1

S

k1 1 r

D

a k112r_br1_{· · ·}1

S

k 1 1 k 1 1

D

b k11

Since~a1b!k11 5~_a1_b!k~_a1_b!5 ~_a1 _b!k_a1 ~_a 1_b!k_{b, multiply the right}

side of Equation (7) bya, then byb, and add, combining like terms. It is also helpful to replace~0 k! by~ 0 k11! and (k k ) by (k11 k11

), since all are equal to 1.

~a1 b!k~_a1_b!5

S

k1 1 0

D

a k111

FS

k 0

D

1

S

k 1

DG

a k_b 1

FS

k 1

D

1

S

k 2

DG

a k21_b21_{· · ·} 1

FS

k r 21

D

1

S

k r

DG

a k112r br1 · · · 1

S

k 11 k 11

D

b k11_.

Apply the additive property given in Equation (6) to the expressions in brackets to get Equation (8), as desired. Therefore, by the Principle of Mathematical Induc-tion, Equation (1) is valid for every position integern.

EXERCISES 8.6

Check Your Understanding

Exercises 1–6 True or False. Give reasons. 1. For every positive integern,~3n!!5~3!!~n!!.

2. There are ten terms in the expression of~11x!10_.

3. The middle term of the expansion of (x11 x)

8 is 70.

4. The expansion of~x2_{12x 11!}8 _{is the same as the} expansion of~x 11!16_.

5. ~18!1~82!2~29!50.

6. For every positive integerx,

S

Ïx11 x

D

4

5 x2₁ 1

x4.

Exercises 7–10 Fill in the blank so that the resulting statement is true.

7. After simplifying the expansion of ~x2 ₂1 x!

5_{, the} coefficient ofx4_{is .}

8. In the expansion of

(

Ïx2 1

Ïx

)

6_{, the middle term is} .

9. ~38!2~28!5 .

10. The number of terms in the expansion of

~x2_14x14!12_{is .}

Develop Mastery

Exercises 1–14 Evaluate and simplify. Use Equations (3)–(6). Then verify by calculator.

1. (a) ~9 3! (b) ~96! 2. (a) ~14 3! (b) ~1411! 3. (a) ~8 5! (b) ~83! 4. (a) ~100 98! (b) ~1002 ! 5. (a) ~20 2!1~203! (b) ~213! 6. (a) ~7 3!1~74! (b) ~84! 7. (a) 5 6·~ 10 5! (b) ~106! 8. (a) 9 4·~ 12 3! (b) ~124! 9. (a) ~10 6!·~63! (b) ~107!·~73! 10. (a) ~12 10!·~104! (b) ~85!·~53! 11. (a) 10! 7! (b) 10! 7! 3! 12. (a) 8!12! (b) 10! 13. (a) 6!14! 3! (b) 8!25! 3! 14. (a) 6!23! (b) ~623!!

Exercises 15–18 Calculator Evaluation Use the Tech-nology Tip to evaluate the expression.

15. (a) ~248! (b) ~373!1~375!

16. (a) ~315! (b) ~164!2~1012!

17. (a) ~128!·~203! (b) ~257!4~254!

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484 Chapter 8 Discrete Mathematics Functions on the Set of Natural Numbers

Exercises 19–24 Evaluate and simplify. 19.

S

n n21

D

20.

S

n n22

D

21.

S

n11 n21

D

22. ~n11!! ~n21!! 23.

S

n k 11

D

S

n k

D

24.

S

n11 r

D

S

n r21

D

Exercises 25–30 Binomial Theorem Use the binomial theorem formula to expand the expression, then simplify your result. 25. ~x21!5 _{26. ~x 23y!}4 27.

S

1 x 22y 2

D

4 28.

S

x2 ₁2 x

D

6 29.

S

3x1 1 x2

D

5 30. ~x 21!7

Exercises 31–34 Expansion Use the formula in Equa-tion (2).(a)Write the expansion in sigma form.(b)Expand and simplify. 31. ~22x!5 _32.

S

_2x1 y 2

D

5 33.

S

x2₁2 x

D

5 34. ~x2_22!6

Exercises 35–38 Number of Terms (a) How many terms are there in the expansion of the given expression?

(b) If the answer in(a)is odd, then find the middle term. If it is even, find the two middle terms.

35. ~x2 _23!8 _36.

S

_x2 ₂1

x

D

15

37. ~11Ïx!5 _{38. ~x 12Ïx!}10

Exercises 39– 40 Find the first three terms in the expan-sion of 39.

S

x11 x

D

20 40.

S

x23 x

D

25

Exercises 41– 44 Find Specified Term If the expression is expanded using Equation (1), find the indicated term and simplify. 41.

S

x3₂2 x

D

5 ; third term 42.

S

x 222y

D

12 ; tenth term 43.

S

2x2 y 2

D

10 ; fourth term 44. ~x21 _12x!8_{; fourth term}

Exercises 45–52 Specified Term If the expression is ex-panded and each term is simplified, find the coefficient of the term that contains the given power of x. See Example 6. 45.

S

x3₂2 x

D

4 ;x4 _46.

S

_{2x 2}1 3

D

10 ;x7 47. ~x2_12!11_;x8 _48.

S

_x2₂2 x

D

10 ;x8 49.

S

x3₂1 x

D

15 ;x25 _50.

S

_x2₂3 x

D

12 ;x9 51. ~x2_22x11!3_;x4 _{52. ~x}2_{14x 14!}3_;x2

Exercises 53–60 Solve Equation Find all positive in-tegers n that satisfy the equation.

53. ~2n!!52~n!! 54. ~3n!!5~3!!~n!! 55. 2~n22!!5n! 56. ~3n!!53~n11!! 57.

S

n 3

D

5

S

n 5

D

58.

S

n 3

D

1

S

n 4

D

5

S

8 4

D

59.

S

n 2

D

515 60.

S

n 2

D

528

61. (a) Show that (106)1(107)5(117) by carrying out the following steps. Using Equation (3), express each term of (106)1(107) as a fraction with factorials; then, without expanding, get a common denominator and express the result as a fraction involving factorials. By Equa-tion (3), show that the result is equal to (117). See Exam-ple 3.

(b) Following a pattern similar to that described in part

(a), prove the additivity property for the binomial coefficients

S

n r

D

1

S

n r11

D

5

S

n11 r11

D

.

62. By expanding the left- and right-hand sides, verify that

S

n k11

D

5 n2k k 11·

S

n k

D

. 63. Explore LetSn51 · 1!12 · 2! 1· · ·1 n·n!11. S151 · 1!115252! and S251 · 1!12 · 2!115653!

(a) EvaluateS3,S4, andS5 and look for a pattern. On the basis of your data, guess the value ofS8. Verify your guess by evaluatingS8directly.

(b) Guess a formula forSnand use mathematical induc-tion to prove that your formula is correct.

64. Explore Supposefn~x!5~x 11x!

2n_{and leta} nbe the middle term of the expansion offn~x!.

(a) Finda1,a2,a3, anda4.

(b) Guess a formula for the general terman. Isan5

~2n!!