How to go from 2*4*6 (2n) to (2^n)n

  • Context: Undergrad 
  • Thread starter Thread starter mathbrah
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Discussion Overview

The discussion revolves around the transformation of the product of even numbers, specifically from the expression 2 * 4 * 6 * ... * (2n) to the form (2^n)n. Participants explore the reasoning and steps involved in this transformation, focusing on mathematical reasoning and factorization.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how to derive the expression (2^n)n from the product of even numbers.
  • Another participant suggests factoring out 2 from the product 2 * 4 * 6 * ... * (2n) as a potential approach.
  • A different participant inquires about the number of factors present in the product 2 * 4 * 6 * ... * (2n).
  • One participant provides a more explicit formulation of the product, rewriting it as (2 * 1) * (2 * 2) * (2 * 3) * ... * (2n).

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the transformation process, and multiple viewpoints regarding the approach remain present.

Contextual Notes

Some assumptions about the properties of products and factorization may be implicit, and the discussion does not resolve how to handle the transition between the two expressions fully.

mathbrah
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how would one see the first

and come up with the second
 
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Have you tried factoring 2 from 2*4*6*...(2n)?
 
In the product 2 * 4 * 6 * ... * (2n), how many factors are explicitly shown there?
 
I write it even a bit more explicitly
[tex]2 \cdot 4 \cdot 6 \cdots (2n)=(2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdots (2n).[/tex]
 

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