Proving Double Factorial: (2n)!=2^nn!

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Discussion Overview

The discussion revolves around proving the identity for double factorials, specifically that (2n)! = 2^n n!. Participants explore different methods of proof, including induction and direct computation, while providing examples to illustrate the identity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests proving the identity by induction, noting that it holds for n = 1 and proposing to show it for n + 1 based on the assumption for general n.
  • Another participant argues that induction is unnecessary and that a direct computation using the definition of the double factorial suffices.
  • A participant provides a breakdown of (2n)! into its factors, highlighting how each even number can be expressed as 2 times an integer and prompting others to consider the implications of this factorization.
  • One participant expresses gratitude for the discussion, indicating engagement with the proposed ideas.

Areas of Agreement / Disagreement

Participants present differing views on the method of proof, with some advocating for induction and others favoring direct computation. No consensus is reached on the preferred approach.

Contextual Notes

Some assumptions about the properties of factorials and double factorials are implicit in the discussion. The steps for the direct computation method are not fully detailed, leaving some mathematical steps unresolved.

Who May Find This Useful

Readers interested in combinatorial mathematics, factorial identities, or methods of mathematical proof may find this discussion relevant.

matematikuvol
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How to prove

(2n)!=2^nn!

for example

(2)!=2^11!=2

(4)!=2^22!=8

...

(2n)!=2n(2n-2)(2n-4)...=2^3n(n-1)(n-2)...

I see that by intiution but I don't know how to write prove.
 
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Prove it by induction. You've shown that it holds for the case n =1, so if you assume that it holds for general n, show that it follows that it holds for the case n+1.
 
You don't need induction. A direct computation using the definition of the double factorial is enough.
 
How?
 
<br /> (2 n)! = 2 \times 4 \times \ldots \times (2 n)<br />

But, notice that:
<br /> \begin{array}{l}<br /> 2 = 2 \times 1 \\<br /> <br /> 4 = 2 \times 2 \\<br /> <br /> \ldots \\<br /> <br /> 2 n = 2 \times n<br /> \end{array}<br />

Combine the first factor of two from each factor in the double factoriel. How many of them are there? What do the remaining factors give?
 
Tnx :)
 

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