(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-(2-1) = 2(k+1)+1 − 1 - Your Proof: Put chosen statements in order in this column and press the Submit Answers button. 1+2+22 Then ·+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 ·+2)+2k+1. (1+2+22++ Thus P(k+1) is true. Thus P(k) is true for all k. = Observe that 120+1 - 1. So P (0) is true. Then 12 +2²+(k+1)² = k(k + 1)(2k + 1) the inductive hypothesis. 6 Now assume that P(k + 1) is true for all k. Note that 1+2 2¹+1 - 1 = 3. = Then 1+2+22+...+2k = 2k+1 -1. Let P(n) be the predicate, "1+2+2² + +22+11". = - by

(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-(2-1) = 2(k+1)+1 − 1 - Your Proof: Put chosen statements in order in this column and press the Submit Answers button. 1+2+22 Then ·+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 ·+2)+2k+1. (1+2+22++ Thus P(k+1) is true. Thus P(k) is true for all k. = Observe that 120+1 - 1. So P (0) is true. Then 12 +2²+(k+1)² = k(k + 1)(2k + 1) the inductive hypothesis. 6 Now assume that P(k + 1) is true for all k. Note that 1+2 2¹+1 - 1 = 3. = Then 1+2+22+...+2k = 2k+1 -1. Let P(n) be the predicate, "1+2+2² + +22+11". = - by

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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