(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: For all n≥0, 1+2+22+ +2=2n+1-1. The correct proof will use 8 of the statements below. Statements to choose from: Now assume that P(k) is true for an arbitrary integer k ≥ 0. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 0. Then 1+2+22 ·+2k+1 =2 (1+2+2² + ... +2k) =2(2k+1-1) =2(k+1)+1-1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Then 1+2+22+ +2k+1 =2k+1 - 1+2k+1 = 2.2k+1-1 =2k+2 - 1 =2(k+1)+1-1 Note that 1+2+22+.... +2k+1 (1+2+2+2)+2k+1. Thus P(k+1) is true. Thus P(k) is true for all k. = Observe that 1=20+1 - 1. So P(0) is true. Then k(k+1)(2k+1) 12+22+1 +(k+1)²= by 6 the inductive hypothesis. Now assume that P(k + 1) is true for all k. Note that 1+2=21+1 - 1 = 3. Then 1+2+22+...+2 = 2k+1 - 1. Let P(n) be the predicate, *1+2+2² + +2=2n+1-1".

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: For all n≥0, 1+2+22+ +2=2n+1-1. The correct proof will use 8 of the statements below. Statements to choose from: Now assume that P(k) is true for an arbitrary integer k ≥ 0. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 0. Then 1+2+22 ·+2k+1 =2 (1+2+2² + ... +2k) =2(2k+1-1) =2(k+1)+1-1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Then 1+2+22+ +2k+1 =2k+1 - 1+2k+1 = 2.2k+1-1 =2k+2 - 1 =2(k+1)+1-1 Note that 1+2+22+.... +2k+1 (1+2+2+2)+2k+1. Thus P(k+1) is true. Thus P(k) is true for all k. = Observe that 1=20+1 - 1. So P(0) is true. Then k(k+1)(2k+1) 12+22+1 +(k+1)²= by 6 the inductive hypothesis. Now assume that P(k + 1) is true for all k. Note that 1+2=21+1 - 1 = 3. Then 1+2+22+...+2 = 2k+1 - 1. Let P(n) be the predicate, *1+2+2² + +2=2n+1-1".

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter8: Sequences, Series, And Probability

Section8.5: Mathematical Induction

Problem 43E

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