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Homework Help: General relation for a pattern?

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data
    hello all,

    I'm in the middle of solving a d.e using the series method. I have come across a weird pattern in part of my solution that I'm confused about:

    6, (6)(10),(6)(10)(14),(6)(10)(14)(18),.....

    2. Relevant equations

    3. The attempt at a solution
    I can see its 2(3), 2(3)*2(5), 2(3)*2(5)*2(7)........ But I'm a little confused on what the nth term is. Any help?

    Also I can see if we start at n=2, (4n-2),(4n-2)(4n-6),(4n-2)(4n-6)(4n-10),.... But does any one know the general term? Thanks!
  2. jcsd
  3. Apr 26, 2015 #2


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    You could write a general term with the double factorial notation. http://en.wikipedia.org/wiki/Double_factorial
  4. Apr 26, 2015 #3


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    You don't really need to use a special notation like the "double factorial".

    I presume you see that it is [(2)(3)][(2)(5)][(2)(7)]...[(2)(2n+1)]. So, first, we have n "2"s : [itex]2^n[/itex].

    Then we have (3)(5)(7)(9)...(2n+1), the product of n consecutive odd numbers. We can write that as
    The numerator is (2n+1)!. We can write the denominator as [itex][2(1)][2(2)][2(3)][2(4)]...[2(n)]= 2^n n![/itex].
  5. Apr 26, 2015 #4
    Thanks for the response hallsofivy,

    I'm a little confused: the pattern is 6 , (6)(10) , (6)(10)(14) , (6)(10)(14)(18) , ....

    And yes it is also 2(3) , 2(3)*2(5) , 2(3)*2(5)*2(7) ect..... But writing the General term as (2n+1)!/2^n(n)! Doesn't fit the pattern.......
  6. Apr 26, 2015 #5


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    You are missing the [itex]2^n[/itex] in the numerator. You should have
    [tex]2^n\frac{(2n+1)!}{2^n n!}= \frac{(2n+1)!}{n!}[/tex]
  7. Apr 26, 2015 #6
    Ahhh! It's so easy! Thanks, for some reason I just couldn't get it...thanks a lot! It's been a while since I've thought about infinite series
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