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Self studying, similar to a Factorial?

  • Thread starter rocomath
  • Start date
  • #1
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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?
 
Last edited:

Answers and Replies

  • #2
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Why is it jumping by 5 every time?
 
  • #3
tiny-tim
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5 x 10 x 15 ...
Hi (rocomath)! :smile:

That would be (5^n) n! :smile:
 
  • #4
1,752
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Hi (rocomath)! :smile:

That would be (5^n) n! :smile:
Argh, I need more practice!!! Thanks tiny-tim :)
 
  • #5
Defennder
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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.
 
  • #6
tiny-tim
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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)! :smile:
:biggrin: I think you're both right! :biggrin:
 
  • #7
Gib Z
Homework Helper
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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, hes 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.
 

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