By dragging statements from the left column to the right column below, give a proof by induction of the following statement: For all n≥2, (1−(1/2^2))(1−(1/3^2))(1−(1/4^2))⋯(1−(1/n^2))=n+(1/2n) Statements to choose from:Thus P(k+ 1) is true.Note that 1Observe that 1-k+12k1+=Let P(n) be the statement,*(¹ − ¹) (¹ - ¹) (¹ - ¹) ··· (¹ – 3)n+1,2n+true.Now assume that P(k) is true for an arbitraryinteger k > 2.Therefore, by the Principle of MathematicalInduction, P(n) is true for all n ≥ 2.2(k+ 1)1²So(¹₁ - 12/²) (¹ - ² - ) --- (¹ - ( ^ + 1)²1111(k+1k+12k+ 12k(k+ 1)²k + 2k² + 2kk + 12k(k+ 1)²k+ 1+k(k + 2)1+12.1(k+ 1)²(k+ 1)²k(k + 2)2+12.2So P(2) isNote that(1(¹ - ) (¹ - ¹³ ) ··· (1¹ -1¹²) =(k+1)²[(¹ - - ) (¹ - ;-) --- (¹ - ¦ )] (¹-1)(k+1)²Your Proof: Put chosen statements in order in thiscolumn and press the Submit Answers button. Then the inductive hypothesis tells us that(¹ − ¹) (¹ − ¹ ) ( ¹ − ¹¦ ) · · · (¹ – 71 – 1111k+22(k+1)*Then(¹ − ²) (¹ − ²¹ ) (¹ − ¹ ) ··· (¹ – ²/² )1111-k+12k==Now assume that P(n) is true for all n ≥ 2.(₁1=11(¹ - - / -) (¹ - - - -) ·· ·- (¹ - (x + 1)²)11113²(k+1k+12kk+12kk+12kk + 22(k+1)So(k+ 1)²k² + 2k(k+1)²(k + 1)²k(k+2)(k+ 1)²Thus P(k) is true for all k.

Answered: Image /qna-images/answer/8b06593e-255a-4b6c-b336-b5324d4ae02d.jpg

Thus P(k+ 1) is true. Note that 1-2 Observe that (1 1- ;) = ² · true. + - k+ 2 2(k+1) Let P(n) be the statement, 1- ·(¹ − ²) (¹ − 3 ) (¹ – ¦ ) ··· (¹ - -/-) n+1" 2n So = Now assume that P(k) is true for an arbitrary integer k > 2. 1 + (1 - (k² + 1)² + k² + 2k Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 2. (k+ 1)² k(k+ 2) (k+ 1)² + k+1+k(k+2) 2k(k+ 1)² 1+1 2.1 1 1 (¹ - - 2² ) (¹ - - - -) ··· (1¹ —- (2 + 1)²) 3² (k+ k+1 2k k+1 2k k+1 2k [(¹-) (¹) [(₁ 2+1 2.2 . So P(2) is (¹ - (1 Note that (1 - 1) (¹ - - - ) --- (1 - (+ 1)² 3² (k+1)² )] (¹- (k+1) ²

Thus P(k+ 1) is true. Note that 1-2 Observe that (1 1- ;) = ² · true. + - k+ 2 2(k+1) Let P(n) be the statement, 1- ·(¹ − ²) (¹ − 3 ) (¹ – ¦ ) ··· (¹ - -/-) n+1" 2n So = Now assume that P(k) is true for an arbitrary integer k > 2. 1 + (1 - (k² + 1)² + k² + 2k Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 2. (k+ 1)² k(k+ 2) (k+ 1)² + k+1+k(k+2) 2k(k+ 1)² 1+1 2.1 1 1 (¹ - - 2² ) (¹ - - - -) ··· (1¹ —- (2 + 1)²) 3² (k+ k+1 2k k+1 2k k+1 2k [(¹-) (¹) [(₁ 2+1 2.2 . So P(2) is (¹ - (1 Note that (1 - 1) (¹ - - - ) --- (1 - (+ 1)² 3² (k+1)² )] (¹- (k+1) ²

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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