8.6 the BINOMIAL THEOREM We Remake Nature by the Act of Discovery, in the Poem Or in the Theorem

Home , Binomial coefficient, Binomial theorem

pg476 [V] G2 5-36058 / HCG / Cannon & Elich cr 11-30-95 MP1

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

(b) Based on your results for (a), guess the minimum number of handshakes. See Exercise 19, page 469. number of moves required if you start with an arbi- Show trary number of n disks. (Hint: Tohelpseeapat- n͑n Ϫ 1͒ tern,add1tothenumberofmovesfornϭ1, 2, 3, f ͑n͒ ϭ for every positive integer n. 4.) 2 (c) A legend claims that monks in a remote monastery 56. Suppose n is an odd positive integer not divisible by 3. areworkingtomoveasetof64disks,andthatthe Show that n 2 Ϫ 1 is divisible by 24. (Hint: Consider the world will end when they complete their sacred three consecutive integers n Ϫ 1, n, n ϩ 1. Explain task. Moving one disk per second without error, 24 why the product ͑n Ϫ 1͒͑n ϩ 1͒ must be divisible by 3 hours a day, 365 days a year, how long would it take andby8.) to move all 64 disks? 57. Finding Patterns If an ϭ ͙24n ϩ 1, (a) write out 55. Number of Handshakes Suppose there are n people the ﬁrst ﬁve terms of the sequence ͕an͖. (b) What odd at a party and that each person shakes hands with every integers occur in ͕an͖? (c) Explain why ͕an͖ contains all other person exactly once. Let f͑n͒ denote the total primes greater than 3. (Hint: UseExercise56.)

8.6 THE BINOMIAL THEOREM 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. ͓And͔ 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 ͑a ϩ b͒n for any positive integer power n,ortofindanyparticularterminsuchanexpansion. We begin by calculating the first few powers directly and then look for significant patterns. To go from one power of ͑a ϩ b͒ to the next, we simply multiply by ͑a ϩ b͒:

͑a ϩ b͒1 ϭ a ϩ b ͑a ϩ b͒2 ϭ a2 ϩ 2ab ϩ b 2 ͑a ϩ b͒3 ϭ a3 ϩ 3a2b ϩ 3ab 2 ϩ b 3

͑ϫ͒ a ϩ b a4 ϩ 3a3b ϩ 3a2b 2 ϩ ab 3 3 2 2 3 4 The thing that started it a b ϩ 3a b ϩ 3ab ϩ b Add like terms all was this silly newspaper puzzle that asked you to count up the total number ͑a ϩ b͒4 ϭ a 4 ϩ 4a 3b ϩ 6a 2b 2 ϩ 4ab 3 ϩ b 4 of ways you could spell the words “Pyramid of Values” ͑ϫ͒ a ϩ b from a triangular array of letters. This led my friend a5 ϩ 4a4b ϩ 6a3b 2 ϩ 4a 2b 3 ϩ ab 4 and me to discover Pascal’s a4b ϩ 4a3b 2 ϩ 6a2b 3 ϩ 4ab 4 ϩ b 5 Add like terms triangle. This happened in grade 10 or 11. Bill Gosper ͑a ϩ b͒5 ϭ a 5 ϩ 5a 4b ϩ 10a3b 2 ϩ 10a2b 3 ϩ 5ab 4 ϩ b 5 pg477 [R] G1 5-36058 / HCG / Cannon & Elich kr 11-20-95 QC1

8.6 The Binomial Theorem 477

When we look at these expansions of ͑a ϩ b͒n for n ϭ 1, 2, 3, 4, and 5, several patterns become apparent. 1. There are n ϩ 1 terms, from an to b n. 2. Every term has essentially the same form: some coefficient times the product of a power of a times a power of b. 3. In each term the sum of the exponents on a and b is always n. 4. The powers (exponents) on a decrease, term by term, from n down to 0 where the last term is given by b n ϭ a0b n, and the exponents on b in- crease from 0 to n. Knowing the form of the terms in the expansion and that the sum of the powers is always n, 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: 11 121 1331 14641 15101051 Using the address notation for Pascal’s triangle that we introduced in Section 5 5 5 5 5 5 8.4, the last row of coefficients in the triangle is ͑ 0͒, ͑ 1 ͒, ͑ 2͒, ͑ 3͒, ͑ 4͒, ͑ 5͒,andso we may write the expansion for ͑a ϩ b͒5: 5 5 5 5 ͑a ϩ b͒5 ϭ a5b 0 ϩ a4b 1 ϩ a3b 2 ϩ a2b 3 ͩ0ͪ ͩ1ͪ ͩ2ͪ ͩ3ͪ 5 5 ϩ a1b 4 ϩ a0b 5 ͩ4ͪ ͩ5ͪ

5 Each term exhibits the same form. For n ϭ 5, each coefﬁcient has the form ͑r͒, where r is also the exponent on b. For each term the sum of the exponents on a and b is always 5, so that when we have b r, we must also have a 5Ϫr.Finally,sincethe ﬁrst term has r ϭ 0, the second term has r ϭ 1, etc., the ͑r ϩ 1͒st term involves r. This leads to a general conjecture for the expansion of ͑a ϩ b͒n which we state as a theorem that can be proved using mathematical induction. (See the end of this section.) Binomial theorem Suppose n is any positive integer. The expansion of ͑a ϩ b͒n is given by n n n n ͑a ϩ b͒n ϭ a nb 0 ϩ a nϪ1b 1 ϩ ···ϩ anϪrbr ϩ···ϩ a0bn ͩ0ͪ ͩ1ͪ ͩrͪ ͩnͪ (1) n where the ͑r ϩ 1͒st term is a nϪrb r,0ՅrՅn. In summation notation, ͩr ͪ n n ͑a ϩ b͒n ϭ anϪrb r. (2) r͚ϭ0 ͩrͪ pg478 [V] G2 5-36058 / HCG / Cannon & Elich kr 11-20-95 QC1

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 of n, and we have not completely justified our use of the name Pascal’s triangle of binomial coefficients. Nonetheless, the multiplication of ͑a ϩ b͒4 by ͑a ϩ b͒ to get the expansion for ͑a ϩ b͒5 contains all the essential ideas of the proof. We still lack a closed-form formula for the binomial coefficients. We know, for 20 20 17 3 example, that the fourth term of the expansion of ͑x ϩ 2y͒ is ͑ 3 ͒x ͑2y͒ ,butwe 20 cannot complete the calculation without the binomial coefficient ͑ 3 ͒.Thiswould 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 n given address within the triangle. He observed that to find ͑r͒,wecantakethe product of all the numbers from 1 through r,anddivideitintotheproductofthe same number of integers, from n downward. This leads to the following formula. Pascal’s formula for binomial coefficients Suppose n is a positive integer and r is an integer that satisfies 0 Ͻ r Յ n. n The binomial coefficient is given by ͩ rͪ n n͑n Ϫ 1͒ ···͑nϪrϩ1͒ ϭ (3) ͩrͪ 1·2·3···r

We leave it to the reader to verify that the last factor in the numerator, ͑n Ϫ r ϩ 1͒,istherth number counting down from n. This gives the same number of factors in the numerator as in the denominator.

᭤EXAMPLE 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 ͑a ϩ b͒10. Solution 10 Strategy: We know ͑ 0 ͒ ϭ Follow the strategy. 1. Use Equation (3) to get 10 10 10 10·9 10 10·9·8 the remaining coefficients. ϭ ϭ 10, ϭ ϭ45, ϭ ϭ120, and ͩ 1 ͪ 1 ͩ 2 ͪ 1·2 ͩ 3 ͪ 1·2·3 10 10·9·8·7 ϭ ϭ210. ͩ 4 ͪ 1·2·3·4 Therefore the first five terms in the expansions of ͑a ϩ b͒10 are a10 ϩ 10a9b ϩ 45a8b 2 ϩ 120a7b 3 ϩ 210a6b 4. ᭣

There is another very common formula for binomial coefﬁcients that uses factorials. 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 n Ϫ r downto1: n n͑nϪ1͒...͑nϪrϩ1͒ ϭ ͩrͪ 1·2·...r n͑nϪ1͒...͑nϪrϩ1͒ ͑nϪr͒...2·1 n! ϭ · ϭ . r! ͑nϪr͒...2·1 r!͑nϪr͒! pg479 [R] G5 5-36058 / HCG / Cannon & Elich kr 12-4-95 OT1

8.6 The Binomial Theorem 479

HISTORICAL NOTE BLAISE PASCAL Pascal’s triangle is named after At the age of 29, Pascal had a Blaise Pascal, born in France in conversion experience that led to a 1623. Pascal was an individual of vow to renounced mathematics for incredible talent and breadth who a life of religious contemplation. made basic contributions in many Before that time, however, in areas of mathematics, but who died addition to his foundational work in early after spending much of life geometry, he built a mechanical embroiled in bitter philosophical computing machine (in honor of and religious wrangling. which the structured computer For some reason, Pascal’s language Pascal is named), father decided that his son should explored relations among binomial not be exposed to any mathematics. coefficients so thoroughly that we All mathematics books in the home Blaise Pascal made call the array of binomial were locked up and the subject was significant contributions to coefficients Pascal’s triangle even banned from discussion. We do not the study of mathematics though the array had been known, know if the appeal of the forbidden before deciding to devote at least in part, several hundred was at work, but young Pascal his life to religion. years earlier, proved the binomial approached his father directly and asked theorem, gave the first published proof by what geometry was. His father’s answer so mathematical induction, and invented (with fascinated the 12-year-old boy that be began Fermat) the science of combinatorial analysis, exploring geometric relationships on his own. He probability, and mathematical statistics. apparently rediscovered much of Euclid Before his death ten years later, Pascal spent completely on his own. When Pascal was only a few days on mathematics. During a night introduced to conic sections ͑see Chapter 10͒ he made sleepless by a toothache, he concentrated on quickly absorbed everything available; he some problems about the cycloid curve that had submitted a paper on conic sections to the attracted many mathematicians of the period. The French academy when he was only 16 years pain subsided, and, in gratitude, Pascal wrote up of age. his work for posterity.

n While Equation (3) does not give a formula for ( 0), the formulation in terms of factorials does apply. n n! n! n! ϭ ϭ ϭ ϭ1. ͩ0ͪ 0! ͑n Ϫ 0͒! 1·n! n! This gives an alternative to Pascal’s formula. Alternative formula for binomial coefficients Suppose n is a positive integer and r an integer that satisfies 0 Յ r Յ n.The n binomial coefficient ͑ r ͒ is given by n n! ϭ (4) ͩrͪ r! ͑n Ϫ r͒! pg480 [V] G2 5-36058 / HCG / Cannon & Elich kr 11-20-95 QC1

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

᭤EXAMPLE 2 Symmetry in binomial coefﬁcients Show that 6 6 n n (a) ϭ (b) ϭ . ͩ2ͪ ͩ4ͪ ͩrͪ ͩn Ϫ rͪ Solution Follow the strategy.

Strategy: Use Equation (4) 6 6! 6! 6 6! 6! to evaluate both sides of the (a) ϭ ϭ and ϭ ϭ . given equations to show that ͩ2ͪ 2! ͑6 Ϫ 2͒! 2! 4! ͩ4ͪ 4! ͑6 Ϫ 4͒! 4! 2! the two sides of each equa- n n! tion are equal. (b) ϭ and ͩrͪ r! ͑n Ϫ r͒! n n! n! ϭ ϭ ͩn Ϫ rͪ ͑n Ϫ r͒! ͓n Ϫ ͑n Ϫ r͔͒! ͑n Ϫ r͒! r! Thus

n n ϭ . ᭣ ͩrͪ ͩn Ϫ rͪ

8 8 9 ᭤EXAMPLE 3 Adding binomial coefﬁcients Show that ͑3͒ ϩ ͑4 ͒ ϭ ͑4 ͒. Get a common denominator and add fractions, but do not evaluate any of the factorials or binomial coefﬁcients. Solution 8 8 Use Equation (3) to get ͑3͒ and ͑4͒, get common denominators, then add.

8 8 8·7·6 8·7·6·5 ͑8·7·6͒·4 8·7·6·5 ϩ ϭ ϩ ϭ ϩ ͩ3ͪ ͩ4ͪ 1·2·3 1·2·3·4 ͑1·2·3͒·4 1·2·3·4 8·7·6͑4ϩ5͒ 9·8·7·6 9 ϭ ϭ ϭ . 1·2·3·4 1·2·3·4 ͩ4ͪ Thus

8 8 9 ϩ ϭ . ᭣ ͩ3ͪ ͩ4ͪ ͩ4ͪ

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 n n denominator and then add. In Example 2, we proved that ( r ) ϭ ͑n Ϫ r͒ giving a symmetry property for the nth row of Pascal’s triangle. Example 3 illustrates the essential steps to prove the following additivity property (see Exercise 61):

n n n ϩ 1 ϩ ϭ ͩrͪ ͩr ϩ 1ͪ ͩr ϩ 1ͪ pg481 [R] G1 5-36058 / HCG / Cannon & Elich cr 11-30-95 MP1

8.6 The Binomial Theorem 481

Symmetry and additivity properties Binomial coefﬁcients have the following properties: n n Symmetry ϭ (5) ͩrͪ ͩn Ϫ rͪ n n n ϩ 1 Additivity ϩ ϭ (6) ͩrͪ ͩr ϩ 1ͪ ͩr ϩ 1ͪ

Notice that we used the additivity property from Equation (6) in Section 8.4 to get the ͑n ϩ 1͒st row from the nth row in Pascal’s triangle. This justifies our claim that the entries in Pascal’s triangle are binomial coefficients. TECHNOLOGY TIP ᭜ Evaluating binomial coefficients

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 notation nCr (meaning “number of combinations taken r at a time,” language from probability), and the key is located in a probability 20 (PRB, PROB) submenu under the MATH menu. To evaluate, say ͑ 4 ͒, the process is as follows.

All TI-calculators: Having entered 20 on your screen, press MATH PRB nCr, which puts nCr on the screen. Then type 4 so you have 20nCr4.Whenyouenter,the display should read 4845. All Casio calculators: Having entered 20 on your screen, press MATH PRB nCr, which puts ރ on the screen. Then type 4 so you have 20ރ4.Whenyou execute, the display should read 4845. HP-38: Press MATH, go down to PROB, highlight COMB, OK. Then, on the command line you want COMB(20,4). Enter to evaluate. HP-48: Put 20 and 4 on the stack. As with many HP-48 operations, COMB (for “combinations”) works with two numbers. MTH NXT PROB COMB returns 4845.

᭤EXAMPLE 4 Binomial Theorem Use the binomial theorem to write out the first five terms of the binomial expansion of ͑x ϩ 2y 2͒20 and simplify. Solution Use Equation (1) with a ϭ x, b ϭ 2y 2,andnϭ20. The first five terms of ͑x ϩ 2y 2͒20 are 20 20 20 20 x20 ϩ x 19͑2y2͒ ϩ x 18͑2y2͒2 ϩ x 17͑2y2͒3 ϩ x 16͑2y2͒4. ͩ 1 ͪ ͩ 2 ͪ ͩ 3 ͪ ͩ 4 ͪ Before simplifying, find the binomial coefficients, using either Equation (3) or the Technology Tip. 20 20 20 20·19 ϭ ϭ 20 ϭ ϭ190 ͩ 1 ͪ 1 ͩ 2 ͪ 1·2 20 20·19·18 20 20·19·18·17 ϭ ϭ1140 ϭ ϭ4845 ͩ 3 ͪ 1·2·3 ͩ 4 ͪ 1·2·3·4 Therefore, the first five terms of ͑x ϩ 2y 2͒20 are x20 ϩ 20·2x19y2 ϩ 190·4x18y4 ϩ 1140 · 8x 17y6 ϩ 4845 · 16x 16y8,or x20 ϩ 40x 19y2 ϩ 760x 18y4 ϩ 9120x 17y6 ϩ 77520x 16y8. ᭣ pg482 [V] G2 5-36058 / HCG / Cannon & Elich kr 11-20-95 QC1

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

2 1 10 ᭤EXAMPLE 5 Finding a middle term In the expansion of ͑2x Ϫ x ͒ , ﬁnd the middle term.

Solution There are 10 ϩ 1or11termsintheexpansionofatenthpower,sothemiddleterm is the sixth (ﬁve before and ﬁve after). The sixth term is given by r ϭ 5.

10 1 5 1 5 ͑2x 2͒5 Ϫ ϭ 252͑32x 10͒ Ϫ ϭϪ8064x 5 ͩ 5 ͪ ͩ xͪ ͩ xͪ

The middle term is Ϫ8064x 5. ᭣

2 1 10 ᭤EXAMPLE 6 Finding a specified term In the expansion of (2x Ϫ x ) , find 1 the term whose simplified form involves x . Solution Strategy: First find the Follow the strategy. The general term given in Equation (2) is general term, then simplify. Finally, find the value of r 10 1 r 10 ͑2x 2͒10Ϫr Ϫ ϭ 210Ϫrx20Ϫ2r͑Ϫ1͒rxϪr that gives Ϫ1 as the expo- r x r nent of x. ͩ ͪ ͩ ͪ ͩ ͪ 10 ϭ ͑Ϫ1͒r210Ϫrx20Ϫ3r. ͩ r ͪ

1 Ϫ1 For the term that involves x or x , ﬁnd the value of r for which the exponent on x is Ϫ1: 20 Ϫ 3r ϭϪ1, or r ϭ 7. The desired term is given by

10 1 7 120͑8͒x 6 960 ͑2x 2͒3 Ϫ ϭϪ ϭϪ . ᭣ ͩ 7 ͪ ͩ xͪ x 7 x

Proof of the Binomial Theorem We can use mathematical induction to prove that Equation (1) holds for every positive integer n.

1 1 1 0 1 0 1 (a) For n ϭ 1, Equation (1) is ͑a ϩ b͒ ϭ ͑0͒a b ϩ ͑1͒a b ϭ a ϩ b,soEqua- tion (1) is valid when n is 1.

k k (b) Hypothesis: ͑a ϩ b͒k ϭ ak ϩ akϪ1b ϩ ··· ͩ0ͪ ͩ1ͪ k k ϩ akϪrbr ϩ···ϩ bk (7) ͩrͪ ͩkͪ k ϩ 1 k ϩ 1 Conclusion: ͑a ϩ b͒kϩ1 ϭ akϩ1 ϩ akb ϩ ··· (8) ͩ 0 ͪ ͩ 1 ͪ k ϩ 1 kϩ1 ϩ a kϩ1Ϫrb r ϩ ···ϩ bkϩ1 ͩ r ͪ ͩkϩ1ͪ pg483 [R] G1 5-36058 / HCG / Cannon & Elich jb 12-1-95 MP3

8.6 The Binomial Theorem 483

Since ͑a ϩ b͒kϩ1 ϭ ͑a ϩ b͒k͑a ϩ b͒ ϭ ͑a ϩ b͒ka ϩ ͑a ϩ b͒kb, multiply the right side of Equation (7) by a,thenbyb, and add, combining like terms. It is also helpful k k ϩ 1 k kϩ1 to replace ͑ 0 ͒ by ͑ 0 ͒ and ( k)by(kϩ1), since all are equal to 1. k ϩ 1 k k ͑a ϩ b͒k͑a ϩ b͒ ϭ akϩ1 ϩ ϩ a kb ͩ 0 ͪ ͫͩ0ͪ ͩ1ͪͬ k k ϩ ϩ a kϪ1b 2 ϩ ··· ͫͩ1ͪ ͩ2ͪͬ k k ϩ ϩ a kϩ1Ϫrb r ϩ ··· ͫͩr Ϫ 1ͪ ͩrͪͬ kϩ1 ϩ bkϩ1. ͩkϩ1ͪ 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 integer n.

EXERCISES 8.6 Check Your Understanding Exercises 1–6 True or False. Give reasons. 8 8 1. For every positive integer n, ͑3n͒! ϭ ͑3!͒͑n!͒. 3. (a) ͑5͒ (b) ͑3͒ 10 100 100 2. There are ten terms in the expression of ͑1 ϩ x͒ . 4. (a) ͑ 98 ͒ (b) ͑ 2 ͒ 1 8 5. (a) 20 20 (b) 21 3. The middle term of the expansion of (x ϩ x ) ͑ 2 ͒ ϩ ͑ 3 ͒ ͑ 3 ͒ . 7 7 8 is 70 6. (a) ͑3͒ ϩ ͑4͒ (b) ͑4͒ 4. x 2 ϩ x ϩ 8 5 10 10 The expansion of ͑ 2 1͒ isthesameasthe 7. (a) · ͑ 5 ͒ (b) ͑ 6 ͒ 16 6 expansion of ͑x ϩ 1͒ . 9 12 12 8. (a) · ͑ 3 ͒ (b) ͑ 4 ͒ 8 8 9 4 5. ͑1͒ ϩ ͑2͒ Ϫ ͑2͒ ϭ 0. 10 6 10 7 9. (a) ͑ 6 ͒ · ͑3͒ (b) ͑ 7 ͒ · ͑3͒ 4 1 2 1 12 10 8 5 6. For every positive integer x, ͙x ϩ ϭ x ϩ . 10. (a) ͑10͒ · ͑ 4 ͒ (b) ͑5͒ · ͑3͒ ͩ xͪ x 4 10! 10! Exercises 7–10 Fill in the blank so that the resulting 11. (a) (b) 7! 7! 3! statement is true. 2 1 5 12. (a) 8! ϩ 2! (b) 10! 7. After simplifying the expansion of ͑ x Ϫ x ͒ ,the 4 6! ϩ 4! 8! Ϫ 5! coefﬁcient of x is . 13. (a) (b) 3! 3! 8. In the expansion of ͙x Ϫ 1 6,themiddletermis ( ͙x ) . 14. (a) 6! Ϫ 3! (b) ͑6 Ϫ 3͒! 8 8 9. ͑3͒ Ϫ ͑2͒ ϭ . 10. The number of terms in the expansion of Exercises 15–18 Calculator Evaluation Use the Tech- ͑x 2 ϩ 4x ϩ 4͒12 is . nology Tip to evaluate the expression. 24 37 37 15. (a) ͑ 8 ! (b) ͑ 3 ͒ ϩ ͑ 5 ͒ 31 16 12 Develop Mastery 16. (a) ͑ 5 ! (b) ͑ 4 ͒ Ϫ ͑ 10͒ 12 20 25 25 Exercises 1–14 Evaluate and simplify. Use Equations 17. (a) ͑ 8 ͒ · ͑ 3 ! (b) ͑ 7 ͒ Ϭ ͑ 4 ͒ (3)–(6). Then verify by calculator. 32 31 31 31 18. (a) ͑ 3 ͒ · ͑ 29͒ (b) ͑ 8 ͒ Ϭ ͑ 3 ͒ 9 9 1. (a) ͑3͒ (b) ͑6͒ 14 14 2. (a) ͑ 3 ͒ (b) ͑11͒ pg484 [V] G2 5-36058 / HCG / Cannon & Elich kr 11-20-95 QC1

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

Exercises 19–24 Evaluate and simplify. Exercises 45–52 Specified Term If the expression is ex- n n n ϩ 1 panded and each term is simplified, find the coefficient of 19. 20. 21. the term that contains the given power of x. See Example 6. ͩn Ϫ 1ͪ ͩn Ϫ 2ͪ ͩn Ϫ 1ͪ 2 4 1 10 n n ϩ 1 45. x 3 Ϫ ; x 4 46. 2x Ϫ ; x7 x 3 ͑n ϩ 1͒! ͩk ϩ 1ͪ ͩ r ͪ ͩ ͪ ͩ ͪ 22. 23. 24. 2 10 ͑n Ϫ 1͒! 47. ͑x 2 ϩ 2͒11; x8 48. x 2 Ϫ ; x8 n n ͩ xͪ k r Ϫ 1 ͩ ͪ ͩ ͪ 1 15 3 12 49. x 3 Ϫ ; x25 50. x 2 Ϫ ; x9 Exercises 25–30 Binomial Theorem Use the binomial ͩ xͪ ͩ xͪ theorem formula to expand the expression, then simplify 51. ͑x 2 Ϫ 2x ϩ 1͒3; x 4 52. ͑x 2 ϩ 4x ϩ 4͒3; x 2 your result. Exercises 53–60 Solve Equation Find all positive in- 25. x Ϫ 1 5 26. x Ϫ 3y 4 ͑ ͒ ͑ ͒ tegers n that satisfy the equation. 1 4 2 6 27. Ϫ 2y 2 28. x 2 ϩ 53. ͑2n͒! ϭ 2͑n!͒ 54. ͑3n͒! ϭ ͑3!͒͑n!͒ ͩx ͪ ͩ xͪ 55. 2͑n Ϫ 2͒! ϭ n! 56. ͑3n͒! ϭ 3͑n ϩ 1͒! 1 5 29. 3x ϩ 30. ͑x Ϫ 1͒7 n n n n 8 ͩ x 2ͪ 57. ϭ 58. ϩ ϭ ͩ3ͪ ͩ5ͪ ͩ3ͪ ͩ4ͪ ͩ4ͪ Exercises 31–34 Expansion Use the formula in Equa- n n 59. ϭ 15 60. ϭ 28 tion (2). (a) Write the expansion in sigma form. (b) Expand ͩ2ͪ ͩ2ͪ and simplify. 10 10 11 61. (a) Show that ( 6 ) ϩ ( 7 ) ϭ ( 7 )bycarryingoutthe 5 5 y following steps. Using Equation (3), express each term 31. ͑2 Ϫ x͒ 32. 2x ϩ 10 10 ͩ 2ͪ of ( 6 ) ϩ ( 7 ) as a fraction with factorials; then, without expanding, get a common denominator and express the 2 5 33. x 2 ϩ 34. ͑x 2 Ϫ 2͒6 result as a fraction involving factorials. By Equa- ͩ xͪ 11 tion (3), show that the result is equal to ( 7 ). See Exam- Exercises 35–38 Number of Terms (a) How many ple 3. terms are there in the expansion of the given expression? (b) Following a pattern similar to that described in part (b) If the answer in (a) is odd, then find the middle term. If (a), prove the additivity property for the binomial it is even, find the two middle terms. coefficients n n n ϩ 1 1 15 ϩ ϭ . 35. ͑x 2 Ϫ 3͒8 36. x 2 Ϫ ͩrͪ ͩr ϩ 1ͪ ͩr ϩ 1ͪ ͩ xͪ 62. By expanding the left- and right-hand sides, verify that 37. ͑1 ϩ ͙x͒5 38. ͑x ϩ 2͙x͒10 n n Ϫ k n ϭ · . Exercises 39–40 Find the first three terms in the expan- ͩk ϩ 1ͪ k ϩ 1 ͩkͪ sion of 63. Explore 1 20 3 25 39. x ϩ 40. x Ϫ Let Sn ϭ 1·1!ϩ2·2! ϩ···ϩ n·n!ϩ1. ͩ xͪ ͩ xͪ S1 ϭ 1·1!ϩ1ϭ2ϭ2! and S ϭ 1·1!ϩ2·2!ϩ1ϭ6ϭ3! Exercises 41–44 Find Specified Term If the expression 2 (a) Evaluate S , S ,andS andlookforapattern.On is expanded using Equation (1), find the indicated term and 3 4 5 the basis of your data, guess the value of S . Verify simplify. 8 your guess by evaluating S8 directly. 2 5 41. x 3 Ϫ ;thirdterm (b) Guess a formula for Sn and use mathematical induc- ͩ xͪ tion to prove that your formula is correct. x 12 64. Explore Suppose f ͑x͒ ϭ ͑ x ϩ 1͒2n and let a be the 42. Ϫ 2y ; tenth term n x n ͩ2 ͪ middle term of the expansion of fn͑x͒. y 10 (a) Find a1, a2, a3,anda4. 43. 2x Ϫ ; fourth term (b) Guess a formula for the general term a .Isa ϭ 2 n n ͩ ͪ ͑2n͒! Ϫ1 8 ? 44. ͑x ϩ 2x͒ ; fourth term n!n!