Self studying, similar to a Factorial?

[rocomath]

I don't have much experience in this, and hopefully someone can recommend the right type of book that I need to look thru in order to solve future problems like this.

I have to have a formula for the nth derivative.

I have like in the numerator, but Idk how to express it properly.

5

5 x 10

5 x 10 x 15 ...

So, my formula would be something like: (5n)!

In my book, it says that n! = 1 x 2 x 3 ... (n-1) x n?

 

[Tedjn]

Why is it jumping by 5 every time?

 

[tiny-tim]

5 x 10 x 15 ...

Hi (rocomath)!

That would be (5^n) n!

 

[rocomath]

Hi (rocomath)!

That would be (5^n) n!

Argh, I need more practice! Thanks tiny-tim :)

 

[Defennder]

Actually I think he meant 5(n!) The placement of the parantheses make a lot of difference.

EDIT: Oh, wait, unless you meant you wanted to represent:

5(5x10)(5x10x15)... instead. Then he's right.

 

[tiny-tim]

Actually I think he meant 5(n!) The placement of the parantheses make a lot of difference.

EDIT: Oh, wait, unless you meant you wanted to represent:

5(5x10)(5x10x15)... instead. Then he's right.

Hi (Defennder)!

I think you're both right! ​

 

[Gib Z]

Actually I think he meant 5(n!) The placement of the parantheses make a lot of difference.

EDIT: Oh, wait, unless you meant you wanted to represent:

5(5x10)(5x10x15)... instead. Then he's right.

No, he's right in the first place.

5 * 10 * 15...5n = ( 5*1 x 5*2 x 5*3...5*n) = 5^n (n!)

There is a factor of 5 for every term in the expression.