Difference between 2n and (2n) ?

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Discussion Overview

The discussion revolves around the interpretation and properties of factorials, specifically addressing the expressions (2n+2)! and (2n)!. Participants explore the definitions and calculations related to these factorials, raising questions about their differences and the correctness of certain factorial identities.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question whether (2n+2)! can be expressed as (2n+2)*(2n+1)*(2n)! or if it should be (2n+2)*(2n)!. There is uncertainty about the correct formulation.
  • There are claims that 2n! is simply twice the value of n!, while (2n)! is the factorial of the number 2n, leading to a distinction between the two expressions.
  • One participant attempts to clarify the factorial definition by stating that n! is the product of all positive integers up to n, and applies this to (2n+2)!. However, there is confusion regarding the manipulation of terms in the factorial expressions.
  • Another participant suggests that the confusion may stem from mixing up the factorial with techniques from mathematical induction, providing an example with n=5 to illustrate their point.

Areas of Agreement / Disagreement

Participants express differing views on the correct interpretation of factorial expressions, indicating that multiple competing perspectives remain without a clear consensus on the matter.

Contextual Notes

Some participants express uncertainty about the definitions and calculations involved in factorials, and there are unresolved questions about the manipulation of factorial terms.

Nikitin
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hey. I've had no education in factorials specifically, but my professor is expecting us to already know this stuff...

In a problem where factorials are included, it is claimed (2n+2)! = (2n+2)*(2n+1)*(2n)!. Shouldn't it be (2n+2)!= (2n+2)*2n! ?

In addition, is there any difference between 2n! and (2n)! ?
 
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1. "(2n+2)!= (2n+2)*2n! ?" No, why?

2. "In addition, is there any difference between 2n! and (2n)! ?"

2*n! is twice the value of n!
(2n!) is the factorial up to the number 2n

For example, 2*3! =12, whereas (2*3)!=6!=720
 
Nikitin said:
hey. I've had no education in factorials specifically, but my professor is expecting us to already know this stuff...

In a problem where factorials are included, it is claimed (2n+2)! = (2n+2)*(2n+1)*(2n)!. Shouldn't it be (2n+2)!= (2n+2)*2n! ?

In addition, is there any difference between 2n! and (2n)! ?

What is a factorial? n! is defined as being all the positive integers up to and including n being multiplied together, so n! = n(n-1)(n-2)...3*2*1

Ok, using this idea, what would (2n+2)! be? Well, first we multiply by (2n+2), then we reduce the value by 1 and multiply by that, so we multiply by ((2n+2)-1) = (2n+1).

Yes,
2n! = 2*(n!) = 2*(n(n-1)(n-2)...3*2*1)

while
(2n)! = (2n)(2n-1)(2n-2)...*3*2*1
 
arildno said:
1. "(2n+2)!= (2n+2)*2n! ?" No, why?

2. "In addition, is there any difference between 2n! and (2n)! ?"

2*n! is twice the value of n!
(2n!) is the factorial up to the number 2n

For example, 2*3! =12, whereas (2*3)!=6!=720

1. Doesn't (2n+2)! = (2n+2)*(2(n-1)+2)*(2(n-2)+2)!

2. Ah yes. How could I miss that? damn, i guess I'm exhausted.

What is a factorial? n! is defined as being all the positive integers up to and including n being multiplied together, so n! = n(n-1)(n-2)...3*2*1

Ok, using this idea, what would (2n+2)! be? Well, first we multiply by (2n+2), then we reduce the value by 1 and multiply by that, so we multiply by ((2n+2)-1) = (2n+1).

Yes,
2n! = 2*(n!) = 2*(n(n-1)(n-2)...3*2*1)

while
(2n)! = (2n)(2n-1)(2n-2)...*3*2*1

ah, OK. thanks 4 the help
 
Nikitin said:
1. Doesn't (2n+2)! = (2n+2)*(2(n-1)+2)*(2(n-2)+2)!

I have a feeling you might be getting this mixed up with the techniques you learned in Mathematical induction?

Let's give n a value, say, n=5

(2n+2)! = (2*5+2)! = 12! = 12*11*...*3*2

Now, 2(n-1)+2 = 2*4+2 = 10.
Notice how 10 is 2 less than 12, because we didn't take 1 away from the value 12, we took a value away from n, which is being multiplied by 2, so if we followed what you wrote we'd end up with 12! = 12*10*8*6*4*2

Essentially, if we have a linear equation an+b for some constants a and b, then [itex]an+b-1 \neq a(n-1)+b[/itex] unless a = 1.
 

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