Understanding the Equation (2n-1)! = (2n-1)(2n)(2n-1)!

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Homework Help Overview

The discussion revolves around the equation involving factorials, specifically questioning the validity of the expression (2n-1)! = (2n-1)(2n)(2n-1)!. Participants are exploring the definitions and properties of factorials in the context of this equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to clarify the original equation and suggest that it may be a misunderstanding of factorial definitions. Others propose alternative factorial expressions and question the validity of the original claim.

Discussion Status

The discussion is active, with participants providing insights into factorial definitions and exploring potential corrections to the original equation. There is no explicit consensus, but several lines of reasoning are being examined.

Contextual Notes

Participants are addressing potential confusion stemming from the definitions of factorials and the specific terms involved in the equation. There is a suggestion that the original poster may have misquoted or misinterpreted a factorial relationship.

fiziksfun
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Can someone please explain to me why (2n-1)! = (2n-1)(2n)(2n-1)! ?? I'm very confused.
 
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It doesn't. Did this come from a book?
 


Do you mean (2n)!= (2n)(2n-1)! ? If so, it comes from the definition of factorial: (2n)!=(2n)(2n-1)(2n-2)...(1)
and (2n-1)!=(2n-1)(2n-2)...(1) so (2n)(2n-1)!=(2n)(2n-1)(2n-2)...(1)=(2n)!
 


fiziksfun said:
Can someone please explain to me why (2n-1)! = (2n-1)(2n)(2n-1)! ?? I'm very confused.
Yes, you are! Dividing both sides of your formula by (2n-1)! you get 1= (2n-1)(2n) which is NOT true!

Perhaps you mean (2n+1)!= (2n+1)(2n)(2n-1)!. That's true because, by definition, (2n+1)!= (2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)...(3)(2)(1). And (2n-1)!= (2n-1)(2n-3)(2n-4)...(3)(2)(1), the "tail end" of that first product. so (2n+1)!= (2n+1)(2n)(2n-1)!
 

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