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

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8.6 the BINOMIAL THEOREM We Remake Nature by the Act of Discovery, in the Poem Or in the 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- n~n 2 1! tern,add1tothenumberofmovesforn51, 2, 3, f ~n! 5 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 2 1 is divisible by 24. (Hint: Consider the world will end when they complete their sacred three consecutive integers n 2 1, n, n 1 1. Explain task. Moving one disk per second without error, 24 why the product ~n 2 1!~n 1 1! 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 5 Ï24n 1 1, (a) write out 55. Number of Handshakes Suppose there are n people the first five terms of the sequence $an%. (b) What odd at a party and that each person shakes hands with every integers occur in $an%? (c) Explain why $an% contains all other person exactly once. Let f~n! 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. @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 ~a 1 b!n 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 1 b! to the next, we simply multiply by ~a 1 b!: ~a 1 b!1 5 a 1 b ~a 1 b!2 5 a2 1 2ab 1 b 2 ~a 1 b!3 5 a3 1 3a2b 1 3ab 2 1 b 3 ~3! a 1 b a4 1 3a3b 1 3a2b 2 1 ab 3 3 2 2 3 4 The thing that started it a b 1 3a b 1 3ab 1 b Add like terms all was this silly newspaper puzzle that asked you to count up the total number ~a 1 b!4 5 a 4 1 4a 3b 1 6a 2b 2 1 4ab 3 1 b 4 of ways you could spell the words “Pyramid of Values” ~3! a 1 b from a triangular array of letters. This led my friend a5 1 4a4b 1 6a3b 2 1 4a 2b 3 1 ab 4 and me to discover Pascal’s a4b 1 4a3b 2 1 6a2b 3 1 4ab 4 1 b 5 Add like terms triangle. This happened in grade 10 or 11. Bill Gosper ~a 1 b!5 5 a 5 1 5a 4b 1 10a3b 2 1 10a2b 3 1 5ab 4 1 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 1 b!n for n 5 1, 2, 3, 4, and 5, several patterns become apparent. 1. There are n 1 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 5 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 ~ 0!, ~ 1 !, ~ 2!, ~ 3!, ~ 4!, ~ 5!,andso we may write the expansion for ~a 1 b!5: 5 5 5 5 ~a 1 b!5 5 a5b 0 1 a4b 1 1 a3b 2 1 a2b 3 S0D S1D S2D S3D 5 5 1 a1b 4 1 a0b 5 S4D S5D 5 Each term exhibits the same form. For n 5 5, each coefficient has the form ~r!, 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 52r.Finally,sincethe first term has r 5 0, the second term has r 5 1, etc., the ~r 1 1!st term involves r. This leads to a general conjecture for the expansion of ~a 1 b!n 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 1 b!n is given by n n n n ~a 1 b!n 5 a nb 0 1 a n21b 1 1 ···1 an2rbr 1···1 a0bn S0D S1D SrD SnD (1) n where the ~r 1 1!st term is a n2rb r,0#r#n. In summation notation, Sr D n n ~a 1 b!n 5 an2rb r. (2) ro50 SrD 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 1 b!4 by ~a 1 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 20 20 17 3 example, that the fourth term of the expansion of ~x 1 2y! is ~ 3 !x ~2y! ,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 ~r!,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 , r # n. n The binomial coefficient is given by S rD n n~n 2 1! ···~n2r11! 5 (3) SrD 1·2·3···r We leave it to the reader to verify that the last factor in the numerator, ~n 2 r 1 1!,istherth number counting down from n. 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 ~a 1 b!10. Solution 10 Strategy: We know ~ 0 ! 5 Follow the strategy. 1. Use Equation (3) to get 10 10 10 10·9 10 10·9·8 the remaining coefficients. 5 5 10, 5 545, 5 5120, and S 1 D 1 S 2 D 1·2 S 3 D 1·2·3 10 10·9·8·7 5 5210. S 4 D 1·2·3·4 Therefore the first five terms in the expansions of ~a 1 b!10 are a10 1 10a9b 1 45a8b 2 1 120a7b 3 1 210a6b 4. 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 n 2 r downto1: n n~n21!...~n2r11! 5 SrD 1·2·...r n~n21!...~n2r11! ~n2r!...2·1 n! 5 · 5 . r! ~n2r!...2·1 r!~n2r!! 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.
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