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Self studying, similar to a Factorial?

  1. May 22, 2008 #1
    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: May 22, 2008
  2. jcsd
  3. May 22, 2008 #2
    Why is it jumping by 5 every time?
     
  4. May 22, 2008 #3

    tiny-tim

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    Hi (rocomath)! :smile:

    That would be (5^n) n! :smile:
     
  5. May 22, 2008 #4
    Argh, I need more practice!!! Thanks tiny-tim :)
     
  6. May 22, 2008 #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.
     
  7. May 23, 2008 #6

    tiny-tim

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    Hi (Defennder)! :smile:
    :biggrin: I think you're both right! :biggrin:
     
  8. May 23, 2008 #7

    Gib Z

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    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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