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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6

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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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6
Transcribed Image Text:(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: The correct proof will use 8 of the statements below. Statements to choose from: k(k + 1)(2k + 1) Then 12+22+... +k² = = Thus P(k) is true for all k. Now assume that P (k) is true for an arbitrary Integer k ≥ 1. So 12+22++(k+1)² k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1)+(k+1)² = (k+1)((k+1)+ 1)(2(k + 1) + 1) = Let P(n) be the predicate, 12 +22+ +7 = n(n + 1)(2n+1), 6 Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Note that 12 KLX1+1)(2-1+1) 6 Thus P(k+1) is true. Then k(k+1)(2k+1) 12+2²++ ++(k+1)²= by the inductive hypothesis. Note that 12+22++(k+1)²= (12+22++k²)+(k+1)². Now assume that P(k + 1) is true for all k. Then we see that 12+2++(k+1)² = k(k+1)(2k+1) +(k+1)² k(k+1)(2k+1) 6(k+1)² + 6 6 2k3 + 3k² + k 6k² + 12k +6 = + 6 6 2k3 +9k2+13k+6 6 (k+1)(k+2)(2k+3) 6 (k+1)((k+1)+ 1)(2(k+1)+1) Your Proof: Put chosen statements in order in this column and press the Submit Answers button. For all n ≥ 1, 12+22+...+n² n(n + 1)(2n+1) = 6
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