1085 11.5 Mathematical Induction 1x + y2k # 1x + y2 = x1x + y2k + y1x + y2k Distributive property = c x # xk + k! 1!1k - 12! xky + k! 2!1k - 22! xk-1y2 + g+ k! 1k - 12! 1! x2yk-1 + xyk d + c xk # y + k! 1!1k - 12! xk-1y2 + g+ k! 1k - 12! 1! xyk + y # yk d The first expression in brackets in equation (3) simplifies to Ak + 1 1 B . To see this, note the following. ak + 1 1 b = 1k + 12 1k2 1k - 12 1k - 22 g1 1 # 1k2 1k - 12 1k - 22 g1 = k + 1 Also, k! 1!1k - 12! + 1 = k1k - 12! 11k - 12! + 1 = k + 1. The second expression becomes Ak + 1 2 B , the last A k + 1 k B , and so on. The result of equation (3) is just equation (2) with every k replaced by k + 1. The truth of Sk implies the truth of Sk+1, which completes the proof of the theorem by mathematical induction. Rearrange terms. 1x + y2k+1 = xk+1 + c k! 1!1k - 12! + 1d xky + c k! 2!1k - 22! + k! 1!1k - 12! d xk-1y2 + g + c 1 + k! 1k - 12! 1! d xyk + yk+1 (3) 11.5 Exercises CONCEPT PREVIEW Write out S4 for each of the following, and decide whether it is true or false. 1. Sn: 3 + 6 + 9 + g+ 3n = 3n1n + 12 2 2. Sn: 12 + 22 + 32 + g+ n2 = n1n + 1212n + 12 6 3. Sn: 1 2 + 1 22 + 1 23 + g+ 1 2n = 2n - 1 2n 4. Sn: 6 + 12 + 18 + g+ 6n = 3n2 + 3n 5. Sn: 2n 62n 6. Sn: n! 76n
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