(n+1)=(n+1)n factorial problem

[ectrhoi]

First of all apologize for my english, I'm french and I'll do my best to be understandable.

So my question is about factorials.
how do you manage to say that (n+1)!=(n+1)n! ? I tried to develop this but my brain is just not able to understand how I'm suppose to do. Could someone please show me how to develop them?

I'm actually doing a problem that requires me to "symplify" (is it the word?) (2n)! and (2(n+1))! and I'm stuck there.

thanks for your help.

 

[jbunniii]

Just write out the factorial as a product:

$$n! = (n)(n-1)(n-2)\ldots(1)$$

$$(n+1)! = ?$$

 

[Mark44]

First of all apologize for my english, I'm french and I'll do my best to be understandable.

You are doing very well. Your English is much better than my French, which is almost nonexistent.

So my question is about factorials.
how do you manage to say that (n+1)!=(n+1)n!

(n + 1)! = (n + 1)*n*(n - 1)*(n - 2)*(n - 3)* ... * 4*3*2*1. Since n! = n*(n - 1)*(n - 2)*(n - 3)* ... * 4*3*2*1, it should be evident that](n + 1)! = (n + 1)*n!

? I tried to develop this but my brain is just not able to understand how I'm suppose to do. Could someone please show me how to develop them?

I'm actually doing a problem that requires me to "symplify" (is it the word?) (2n)! and (2(n+1))! and I'm stuck there.

thanks for your help.

The word is "simplify." (2n)! = (2n)(2n-1)(2n-2)(2n-3)...(n+1)(n)(n-1)(n-2)...(3)(2)(1). Note that this is not the same as 2n!, which is 2(n!). Can you continue from here?

 

[ectrhoi]

You are doing very well. Your English is much better than my French, which is almost nonexistent.
(n + 1)! = (n + 1)*n*(n - 1)*(n - 2)*(n - 3)* ... * 4*3*2*1. Since n! = n*(n - 1)*(n - 2)*(n - 3)* ... * 4*3*2*1, it should be evident that](n + 1)! = (n + 1)*n!

The word is "simplify." (2n)! = (2n)(2n-1)(2n-2)(2n-3)...(n+1)(n)(n-1)(n-2)...(3)(2)(1). Note that this is not the same as 2n!, which is 2(n!). Can you continue from here?

From what I understood from the (n+1)!=(n+1)n! example I would develop (2n)! like this: (2n)(n)(n-1)(n-2)...(3)(2)(1) but obviously it isn't the case?

Thank you very much for your help, I'm feeling I'm on the way of understanding!

 

[Mark44]

From what I understood from the (n+1)!=(n+1)n! example I would develop (2n)! like this: (2n)(n)(n-1)(n-2)...(3)(2)(1) but obviously it isn't the case?

Right, that's not the case. The expansion above is omitting all of the factors (2n-1)(2n-2)...(n+1)

Thank you very much for your help, I'm feeling I'm on the way of understanding!