Result For every positive integer n, Proof We proceed by induction. Since We show that (2.1)! 2 2¹.1! 2' the statement is true for n = 1. Assume, for a positive integer k, that 1.3.5... (2k-1)= Observe that (2k + 2)! 2k+1. (k+1)! 1.3.5 (2n − 1) = = 1= 1.3.5...(2k + 1) = (2k + 2)(2k + 1) 2. (k+1) = 1 · 3 · 5 · · · (2k + 1). By the Principle of Mathematical Induction, for every positive integer n. = (2k)! 2k. (k)! (2n)! 2n.n! (2k + 2)! 2k+1. (k+1)! (2k)! 2k. k! 1.3.5... (2n-1) = = (2k + 1)[1 · 3 · 5 · · · (2k − 1)] (2n)! 2n.n!

Result For every positive integer n, Proof We proceed by induction. Since We show that (2.1)! 2 2¹.1! 2' the statement is true for n = 1. Assume, for a positive integer k, that 1.3.5... (2k-1)= Observe that (2k + 2)! 2k+1. (k+1)! 1.3.5 (2n − 1) = = 1= 1.3.5...(2k + 1) = (2k + 2)(2k + 1) 2. (k+1) = 1 · 3 · 5 · · · (2k + 1). By the Principle of Mathematical Induction, for every positive integer n. = (2k)! 2k. (k)! (2n)! 2n.n! (2k + 2)! 2k+1. (k+1)! (2k)! 2k. k! 1.3.5... (2n-1) = = (2k + 1)[1 · 3 · 5 · · · (2k − 1)] (2n)! 2n.n!

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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