1084 CHAPTER 11 Further Topics in Algebra 2 = 21, so add the exponents. EXAMPLE 3 Using the Generalized Principle Let Sn represent the statement 2n 72n + 1. Show that S n is true for all values of n such that n Ú 3. SOLUTION (Check that Sn is false for n = 1 and n = 2.) Step 1 Show that Sn is true for n = 3. 23 72 # 3 + 1 Let n = 3. 8 77 True Thus, S3 is true. Step 2 Now show that Sk implies Sk+1, where k Ú 3, and where Sk is 2k 72k + 1, and S k+1 is 2k+1 721k + 12 + 1. Start with Sk and assume it is a true statement. 2k 72k + 1 2 # 2k 7212k + 12 Multiply each side by 2. 2k+1 74k + 2 Product rule; distributive property 2k+1 72k + 2 + 2k Rewrite 4k as 2k + 2k. 2k+1 721k + 12 + 2k Factor 2k + 2 on the right. Because 2k 71 for positive integers k Ú 3, replacing 2k with 1 will maintain the truth value of this inequality. 2k+1 721k + 12 + 1 S k+1 Thus, Sk implies Sk+1. Together with the fact that S3 is true, this shows that Sn is true for every positive integer value of n greater than or equal to 3. S Now Try Exercise 27. Proof of the Binomial Theorem The binomial theorem can be proved by mathematical induction. 1x + y2n = xn + a n 1b xn-1y + a n 2b xn-2y2 + a n 3b xn-3y3 For any positive integer n and any complex numbers x and y + g+ a n rb xn-ryr + g+ a n n - 1b xyn-1 + yn (1) Proof Let Sn be statement (1). Begin by verifying Sn for n = 1. S1: 1x + y21 = x1 + y1 True Now assume that Sn is true for the positive integer k. Statement Sk becomes Sk: 1x + y2k = xk + k! 1!1k - 12! xk-1y + k! 2!1k - 22! xk-2y2 Definition of the binomial coefficient + g+ k! 1k - 12! 1! xyk-1 + yk. (2) Multiply each side of equation (2) by x + y.
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