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Result For every positive integer n, 1.3.5 (2n-1) = (2n)! 2n.n!

Result For every positive integer n, 1.3.5 (2n-1) = (2n)! 2n.n!

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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Please, if able show the highlighted part in more steps, I understand what is being done until the highlighted part, that goes way too fast for me,

 

Result For every positive integer n, Proof We proceed by induction. Since We show that (2.1)! 2 2¹.1! 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. ¹. 1.3.5... (2n-1) = (k+1)! (2k + 2)(2k + 1) 2. (k+1) = 1.3.5 (2k + 1). By the Principle of Mathematical Induction, = 1= 1.3.5 (2k + 1) = for every positive integer n. = (2n)! 2nn! (2k)! 2k. (k)! (2k + 2)! 2k+1. (k+1)! (2k)! 2k. k! 1.3.5...(2n - 1) = = (2k + 1)[1 · 3 · 5 · · · (2k − 1)] (2n)! 2n.n!
Transcribed Image Text:Result For every positive integer n, Proof We proceed by induction. Since We show that (2.1)! 2 2¹.1! 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. ¹. 1.3.5... (2n-1) = (k+1)! (2k + 2)(2k + 1) 2. (k+1) = 1.3.5 (2k + 1). By the Principle of Mathematical Induction, = 1= 1.3.5 (2k + 1) = for every positive integer n. = (2n)! 2nn! (2k)! 2k. (k)! (2k + 2)! 2k+1. (k+1)! (2k)! 2k. k! 1.3.5...(2n - 1) = = (2k + 1)[1 · 3 · 5 · · · (2k − 1)] (2n)! 2n.n!
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