Exercise 4 Infinite Geometric Progressions The Binomial Theorem. Evaluate each limit. 1. lim (b c 5) 2. lim (a 3 b 3. lim. a b c 4.

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1 Section 5 5 The Binomial Theorem 717 Exercise 4 Infinite Geometric Progressions Evaluate each limit. 1. lim (b c 5) b 0. lim (a b 0 b ) 3 b 3. lim b 0 c 4 a b c 4. lim b 0 b c 5 Find the sum of the infinitely many terms of each GP , 7, 36, 18, , 4, 1, 1 4, ,, 0.4, 0.08, , 1 4, 1 16,... Write each decimal number as a rational fraction Each swing of a certain pendulum is 78% as long as the one before. If the first swing is 0 cm, find the entire distance travelled by the pendulum before it comes to rest. 14. A ball dropped from a height of 10 m rebounds to three-fourths of its height on each bounce. Find the total distance travelled by the ball before it comes to rest. 5 5 The Binomial Theorem Powers of a Binomial Recall that a binomial, such as (a b), is a polynomial with two terms. By actual multiplication we can show that (a b) 1 a b (a b) a ab b (a b) 3 a 3 3a b 3ab b 3 (a b) 4 a 4 4a 3 b 6a b 4ab 3 b 4 We now want a formula for (a b) n with which to expand a binomial without actually carrying out the multiplication. In the expansion of (a b) n, where n is a positive integer, we note the following patterns: 1. There are n 1 terms.. The power of a is n in the first term, decreases by 1 in each later term, and reaches 0 in the last term. 3. The power of b is 0 in the first term, increases by 1 in each later term, and reaches n in the last term. 4. Each term has a total degree of n. (That is, the sum of the degrees of the variables is n.) 5. The first coefficient is 1. Further, the product of the coefficient of any term and its power of a, divided by the number of the term, gives the coefficient of the next term. (This property gives a recursion formula for the coefficients of the binomial expansion.) The formula is expressed as follows: n (n 1) n (n 1) (n ) (a b) n a n na n 1 b a n b a n 3 b 3 b n (3)

2 718 Chapter 5 Sequences, Series, and the Binomial Theorem Example 36: Use the binomial theorem to expand (a b) 5. Solution: From the binomial theorem, (a b) 5 a 5 5a 4 b 5 (4) 5 (4)(3) 5 (4)(3)() a 3 b a b 3 ab 4 b 5 (3) (3)(4) a 5 5a 4 b 10a 3 b 10a b 3 5ab 4 b 5 which can be verified by actual multiplication. If the expression contains only one variable, we can verify the expansion graphically. Example 37: Use the result from Example 36 to expand (x 1) 5, and verify graphically. Solution: Substituting a x and b 1 into the preceding result gives (x 1) 5 x 5 5x 4 10x 3 10x 5x 1 We will use the binomial theorem to obtain the values in a binomial probability distribution in the following chapter on statistics. To verify graphically, we plot the following on the same axes: and y 1 (x 1) 5 y x 5 5x 4 10x 3 10x 5x 1 The two graphs, shown in Fig. 5 6, are identical. This is not a proof that our expansion is correct, but it certainly gives us more confidence in our result. y x 4 6 FIGURE 5 6 Graphs of y (x 1) 5 and y x 5 5x 4 10x 3 10x 5x 1. Factorial Notation We now introduce factorial notation. For a positive integer n, factorial n, written n!, is the product of all of the positive integers less than or equal to n. n! 1 3 n Of course, it does not matter in which order we multiply the numbers. We may also write n! n(n 1) 3 1 Example 38: (a) 3! (b) 5!

3 Section 5 5 The Binomial Theorem 719 (c) 6!3! 7! 6!() (3) 7 (6!) 6 7 (d) 0! 1 by definition Binomial Theorem Written with Factorial Notation The use of factorial notation allows us to write the binomial theorem more compactly. Thus: Binomial Theorem n(n 1) n(n 1) (n ) (a b) n a n na n 1 b a n b a n 3 b 3 b n 45! 3! Example 39: Using the binomial formula, expand (a b) 6. Solution: There will be (n 1) or seven terms. Let us first find the seven coefficients. The first coefficient is always 1, and the second coefficient is always n, or 6. We calculate the remaining coefficients in Table 5 4. TABLE 5 4 Term 1 1 n 6 n (n 1) 3 6 (5) 15! 1 () n (n 1) (n ) 6 (5)(4) 4 0 3! 1 ()(3) Coefficient n (n 1) (n ) (n 3) 6 (5)(4)(3) 15 4! 1 ()(3)(4) n (n 1) (n ) (n 3) (n 4) 6 (5) (4) (3) () 5! 1 () (3) (4) (5) 6 n (n 1) (n ) (n 3) (n 4) (n 5) 6 (5) (4) (3) () (1) 6! 1 () (3) (4) (5) (6) 1 The seven terms thus have the coefficients and our expansion is (a b) 6 a 6 6a 5 b 15a 4 b 0a 3 b 3 15a b 4 6ab 5 b 6 Pascal s Triangle We now have expansions for (a b) n, for values of n from 1 to 6, to which we add (a b) 0 1. Let us now write the coefficients only of the terms of each expansion. Exponent n

4 70 Chapter 5 Sequences, Series, and the Binomial Theorem Pascal s triangle is named for Blaise Pascal (163 6), the French geometer, probabilist, combinatorist, physicist, and philosopher. This array is called Pascal s triangle. We may use it to predict the coefficients of expansions with powers higher than 6 by noting that each number in the triangle is equal to the sum of the two numbers above it (thus the 15 in the lowest row is the sum of the 5 and the 10 above it). When expanding a binomial, we may take the coefficients directly from Pascal s triangle, as shown in the following example. Example 40: Expand (1 x) 6, obtaining the binomial coefficients from Pascal s triangle. Solution: From Pascal s triangle, for n 6, we read the coefficients We now substitute into the binomial formula with n 6, a 1, and b x. When one of the terms of the binomial is negative, as here, we must be careful to include the minus sign whenever that term appears in the expansion. Thus (1 x) 6 [1 ( x)] (1 5 )( x) 15(1 4 )( x) 0(1 3 )( x) 3 15(1 )( x) 4 6(1)( x) 5 ( x) 6 1 6x 15x 0x 3 15x 4 6x 5 x 6 If either term in the binomial is itself a power, a product, or a fraction, it is a good idea to enclose that entire term in parentheses before substituting. Example 41: Expand (3 x y 3 ) 4. Solution: The binomial coefficients are We substitute 3 x for a and y 3 for b. 4 p 3 q y3 p 3 q x [ (y 3 ) x ] p 3 x 81 x 8 q 4 4 p 3 q 3 (y 3 ) 6 p 3 x x 16y3 x 6 16y6 x 4 96y9 x 16y 1 q (y 3 ) 4 p 3 q (y 3 ) 3 (y 3 ) 4 x Sometimes we might not need the entire expansion but only the first several terms. Example 4: Find the first four terms of (x y 3 ) 11. Solution: From the binomial formula, with n 11, a x, and b ( y 3 ), (x y 3 ) 11 [(x ) ( y 3 )] 11 (x ) (x ) 10 ( y 3 ) 11(10) (x ) 9 ( y 3 ) 11 (10) (9) (x ) 8 ( y 3 ) 3 (3) x x 0 y 3 0x 18 y 6 130x 16 y 9 A trinomial can be expanded by the binomial formula by grouping the terms, as in the following example.

5 Section 5 5 The Binomial Theorem 71 Example 43: Expand (1 x x ) 3 by the binomial formula. Solution: Let z x x. Then (1 x x ) 3 (1 z) 3 1 3z 3z z 3 Substituting back, we have (1 x x ) 3 1 3(x x ) 3(x x ) (x x ) 3 1 6x 9x 4x 3 9x 4 6x 5 x 6 General Term We can write the general or rth term of a binomial expansion (a b) n by noting the following pattern: 1. The power of b is 1 less than r. Thus a fifth term would contain b 4, and the rth term would contain b r 1.. The power of a is n minus the power of b. Thus a fifth term would contain a n 5 1, and the rth term would contain a n r The coefficient of the rth term is n! (r 1)!(n r 1)! The formula is expressed as follows: General Term In the expansion for (a b) n, n! 46 r th term (r 1)! (n r 1)! an r 1 b r 1 Example 44: Find the eighth term of (3a b 5 ) 11. Solution: Here, n 11 and r 8. So n r 1 4. Substitution yields n! 11! 330 (r 1)! (n r 1)! 7! 4! Then the eighth term 330(3a ) (b 5 ) (81 a 4 )(b 35 ) 6,730 a 4 b 35 Proof of the Binomial Theorem We have shown that the binomial theorem is true for n 1,, 3, 4, or 5. But what about some other exponent, say, n 50? To show that the theorem works for any positive integral exponent, we give the following proof. We first show that if the binomial theorem is true for some exponent n k, it is also true for n k 1. We assume that (a b) k a k ka k 1 b Pa k r b r Qa k r 1 b r 1 Ra k r b r b k (1) where we show three intermediate terms with coefficients P, Q, and R. Multiplying both sides of Eq. (1) by (a b) gives

6 7 Chapter 5 Sequences, Series, and the Binomial Theorem We have not written the b r term or the b r 3 term because we do not have expressions for their coeffi cients, and they are not needed for the proof. (a b ) k 1 a k 1 ka k b Pa k r 1 b r Qa k r b r 1 Rak r 1 b r ab k a k b Pa k r b r 1 Qa k r 1 b r Ra k r b r 3 b k 1 (a b ) k 1 a k 1 (k 1) a k b (P Q) a k r b r 1 (Q R) a k r 1 b r ab k b k 1 () after combining like terms. Does Eq. () obey the binomial theorem? First, the power of b is 0 in the first term and increases by 1 in each later term, reaching k 1 in the last term, so property 3 is met. Thus there are k 1 terms containing b, plus the first term that does not, making a total of k terms, so property 1 is met. Next, the power of a is k 1 in the first term and decreases by 1 in each succeeding term until it equals 0 in the last term, making the total degree of each term equal to k 1, so properties and 4 are met. Further, the coefficient of the first term is 1, and the coefficient of the second term is k 1, which agrees with property 5. Finally, in a binomial expansion, the product of the coefficient of any term and its power of a, divided by the number of the term, must give the coefficient of the next term. Here, the first of our two intermediate terms has a coefficient (P Q). The coefficient of its a term is k r, the number of that term is r, and the coefficient of the next term is (Q R). Thus the identity k r (P Q) Q R r must be true. But since we must show that P(k r) Q r 1 P(k r) [ P r 1 ] k r r and R P(k r) r 1 Q(k r 1) r P(k r)(k r 1) (r 1)(r ) Working each side separately, we combine each over a common denominator. P(k r)(r 1) (r 1)(r ) P(k r)(k r) (r 1)(r ) P(k r)(r ) (r 1)(r ) P(k r)(k r 1) (r 1)(r ) Since the denominators are the same on both sides, we simply have to show that the numerators are identical. Factoring gives us This type of proof is called mathematical induction. P(k r)(r 1 k r) P(k r)(r k r 1) P(k r)(k 1) P(k r)(k 1) This completes the proof that if the binomial theorem is true for some exponent n k, it is also true for the exponent n k 1. But since we have shown, by direct multiplication, that the binomial theorem is true for n 5, it must be true for n 6. And since it is true for n 6, it must be true for n 7, and so on, for all of the positive integers. Fractional and Negative Exponents We have shown that the binomial formula is valid when the exponent n is a positive integer. We will not prove it here, but the binomial formula also holds when the exponent is negative or fractional. However, we now obtain an infinite series, called a binomial series, with no last term. Further, the binomial series is equal to (a b ) n only if the series converges. Example 45: Write the first four terms of the infinite binomial expansion for 5 x.

7 Section 5 5 The Binomial Theorem 73 Solution: We replace the radical with a fractional exponent and substitute into the binomial formula in the usual way. (5 x) 1/ 5 1/ 1 p q 5 1/ x (1/)( 1/) 5 3/ x (1/)( 1/)( 3/) 5 5/ x 3 6 x x (5) x 3 3/ 5/ 16 (5) Switching to decimal form and working to five decimal places, we get (5 x) 1/ x x x 3 We now ask if this series is valid for any x. Let s try a few values. For x 0, (5 0) 1/ which checks within the number of digits retained. For x 1, (5 1) 1/ which compares with a value for 6 of Not bad, considering that we are using only four terms of an infinite series. To determine the effect of larger values of x, we have used a computer to generate eight terms of the binomial series as listed in Table 5 5. TABLE 5 5 x 5 x by Series 5 x by Calculator Error Into it we have substituted various x values, from 0 to 11, and compared the value obtained by series with that from a calculator. The values are plotted in Fig y y = 5 + x By series 4 By calculator x FIGURE 5 7

8 74 Chapter 5 Sequences, Series, and the Binomial Theorem Notice that the accuracy gets worse as x gets larger, with useless values being obtained when x is 6 or greater. This illustrates the following: A binomial series is equal to (a b) n only if the series converges, and that occurs when a b. Our series in Example 45 thus converges when x 5. The binomial series, then, is given by the following formula: Binomial Series n (n 1) (a b) n a n na n 1 b a n b! n (n 1) (n ) a n 3 b 3 3! where a b 47 The binomial series is often written with a 1 and b x, so we give that form here also. Binomial Series (1 x) n 1 nx n (n 1) n( n 1) (n ) x x 3! 3! where x 1 48 Example 46: Expand to four terms: 1 3 1/a 3b 1/3 Solution: We rewrite the given expression without radicals or fractions, and then expand. 1 3 [(a 1 ) ( 3b 1/3 )] 1/3 1/a 3b 1/3 (a 1 ) 1/3 1 (a 1 ) 4/3 ( 3b 1/3 ) 3 9 (a 1 ) 7/3 ( 3b 1/3 ) 14 (a 1 ) 10/3 ( 3b 1/3 ) 3 81 a 1/3 a 4/3 b 1/3 a 7/3 b / a10/3 b Exercise 5 The Binomial Theorem Factorial Notation Evaluate each factorial. 1. 6!. 8! 3. 7! 5! 4. 11! 9!! 5. 7! 3! 4! 6. 8! 3! 5!

9 Section 5 5 The Binomial Theorem 75 Binomials Raised to an Integral Power Verify each expansion. Obtain the binomial coefficients by formula or from Pascal s triangle as directed by your instructor. 7. (x y) 7 x 7 7x 6 y 1x 5 y 35x 4 y 3 35x 3 y 4 1x y 5 7xy 6 y 7 8. (4 3b) b 864b 43b 3 81b 4 9. (3a b) 4 81a 4 16a 3 b 16a b 96ab 3 16b (x 3 y) 7 x 1 7x 18 y 1x 15 y 35x 1 y 3 35x 9 y 4 1x 6 y 5 7x 3 y 6 y (x 1/ y /3 ) 5 x 5/ 5x y /3 10x 3/ y 4/3 10xy 5x 1/ y 8/3 y 10/3 1. (1 x 1 y ) 3 1 x 3 3 x y 3 xy 4 1 y (a b b a) 6 (a b) 6 6(a b) 4 15(a b) 0 15(b a) 6(b a) 4 (b a) (1 x y ) 4 x 4x 3/ y 6x 1 y 4 4x 1/ y 6 y (a b ) 5 3a 10 80a 8 b 1/ 80a 6 b 40a 4 b 3/ 10a b b 5/ 16. ( x c /3 ) 4 x 4x 3/ c /3 6xc 4/3 4x 1/ c c 8/3 Verify the first four terms of each binomial expansion. 17. (x y 3 ) 8 x 16 8x 14 y 3 8x 1 y 6 56x 10 y (x y 3 ) 11 x x 0 y 3 0x 18 y 6 130x 16 y (a b 4 ) 9 a 9 9a 8 b 4 36a 7 b 8 84a 6 b 1 0. (a 3 1 b) 1 a 36 4a 33 b 64a 30 b 1760a 7 b 3 Trinomials Verify each expansion. 1. (a a 3) 3 8a 6 1a 5 4a 4 37a 3 63a 7a 7. (4x 3xy y ) 3 64x 6 144x 5 y 60x 4 y 45x 3 y 3 15x y 4 9xy 5 y 6 3. (a a 4) 4 16a 8 3a 7 104a 6 184a 5 89a 4 368a 3 416a 56a 56 General Term Write the requested term of each binomial expansion, and simplify. 4. Seventh term of (a b 3 ) 1 5. Eleventh term of ( x) Fourth term of (a 3b) 7 7. Eighth term of (x a) Fifth term of (x y ) 5 9. Ninth term of (x 1) 15 Binomials Raised to a Fractional or Negative Power Verify the first four terms of each infinite binomial series. 30. (1 a) /3 1 a 3 a /9 4a a 1 a a 8 a (1 5a) 5 1 5a 375a 4375a (1 a) 3 1 3a 6a 10a a 1 a 6 7a 7 91a 3 196

10 76 Chapter 5 Sequences, Series, and the Binomial Theorem Case Study Discussion Forever Payments In economics, we find a discussion of present value of money. The idea is that $1,000 a year from now would have a present value of $95.38, assuming that you can invest this amount at 5% interest. That is, $95.38 invested at 5% would grow to $1,000 in one year. How do we calculate this? This formula can calculate the value of any amount accruing interest: A P(1 n). The final amount A is the principle P times 1 interest rate (expressed as a decimal). So if we know the final amount A and want to calculate the principle, we can use the equation P A/(1 i). To earn $1,000 in one interest period at, say, 5% interest, we would need P $1,000/(1 0.05) $ If the payment were two years away it would be 1000/(1 0.05) $ The power of two represents the two interest payments. So, if we receive payments of $1,000 in year one and $1,000 in the second year, the total cost of that $,000 would be $95.38 $ $1, (again assuming an investment of 5% per year, simple interest). As you can see, we can write the present value of the two $1,000 payments as 1000 (1 i) (1 i) But how do you calculate a present value of money if the money keeps going and going forever? If we extend the formula above to infinite payments, the formula becomes a series: P 1000 n =1 (1 i) n This is a geometric series with the common ratio of 1 (1 i). Remember that the present value of the first payment, a, is The sum of the series is (1 i) a 1 r 1000 (1 i) 1 1 (1 i) 1000 i This states that an infinite number of $1,000 payments would have a present value of $1,000/0.05 = $0,000. Now, this actually does make sense because $0,000 invested perpetually at 5% would generate $1,000 worth of interest every year, forever. CHAPTER 5 REVIEW PROBLEMS 1. Find the sum of seven terms of the AP: 4, 1,,.... Find the tenth term of (1 y) Insert four arithmetic means between 3 and How many terms of the AP 5,, 1,... will give a sum of 63? 5. Insert four harmonic means between and Find the twelfth term of an AP with first term 7 and common difference Find the sum of the first 10 terms of the AP 1, 3, 4 1 3,...