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The right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] (k+1)+1 (k+1)+1 We must show that P(k + 1) is true. The left-hand side of P(k + 1) is i· 2'. When the final term of i. 2' is written separately, the result is i- 2 = i• 2i + - The i= 1 right-hand side of P(k + 1) is After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes + (k + 2)2k + When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal + 2. Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It

The right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] (k+1)+1 (k+1)+1 We must show that P(k + 1) is true. The left-hand side of P(k + 1) is i· 2'. When the final term of i. 2' is written separately, the result is i- 2 = i• 2i + - The i= 1 right-hand side of P(k + 1) is After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes + (k + 2)2k + When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal + 2. Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] (k+1)+1 (k+1)+1 (k+1)+1 We must show that P(k + 1) is true. The left-hand side of P(k + 1) is i: 2'. When the final term of 5i· 2' is written separately, the result is Σ i· 2' = Si. 2' + The | = 1 i = 1 i = 1 i = 1 + (k + 2)2* + 2). right-hand side of P(k + 1) is After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal + 2. Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It
Transcribed Image Text:The right-hand side of P(k) is [The inductive hypothesis states that the two sides of P(k) are equal.] (k+1)+1 (k+1)+1 (k+1)+1 We must show that P(k + 1) is true. The left-hand side of P(k + 1) is i: 2'. When the final term of 5i· 2' is written separately, the result is Σ i· 2' = Si. 2' + The | = 1 i = 1 i = 1 i = 1 + (k + 2)2* + 2). right-hand side of P(k + 1) is After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal + 2. Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Need Help? Read It
Prove the following statement by mathematical induction. n + 1 For every integer n 2 0, Fi. 2' = n· 2" + 2 + 2. i = 1 Proof (by mathematical induction): Let P(n) be the equation n + 1 si.2' = n· 2" + 2 + 2. i = 1
Transcribed Image Text:Prove the following statement by mathematical induction. n + 1 For every integer n 2 0, Fi. 2' = n· 2" + 2 + 2. i = 1 Proof (by mathematical induction): Let P(n) be the equation n + 1 si.2' = n· 2" + 2 + 2. i = 1
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