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Thread: Sequence with factorial

  1. #1
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    Sequence with factorial

    I am having problem with this question. Please help

    Find showing all working, a recursive definition of the sequence with general term

    $\displaystyle tn = 2(n+1)!5^n$ from n1 onwards

    How am I suppose to find the recursive definition? What are the steps?
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  2. #2
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    Re: Sequence with factorial

    A recursive definition requires two components
    1) A $\displaystyle t_1$ (a base term)
    2) A general formula relating $\displaystyle t_{n+1}$ and $\displaystyle t_{n}$ for $\displaystyle n \geq 1$ where n is a natural number
    --

    Finding $\displaystyle t_1$ is easy. Replace $\displaystyle n=1$ to get $\displaystyle t_1 = 2(1+1)!5^1 = 20$

    Since we know the formula for $\displaystyle t_n$ we also know the formula for $\displaystyle t_{n+1}$ (replace n with n+1)
    Now consider $\displaystyle \frac{t_{n+1}}{t_n}$, the ratio between two consecutive terms, and isolate $\displaystyle t_{n+1}$ for the recursive formula as desired. I'll leave the rest to you.
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    Re: Sequence with factorial

    Macsters thank you very much for the reply.

    I don't think I came up with the right answer and I can't simply anymore because my algebra is very bad.

    $\displaystyle tn+1/tn = \frac{2(n+2)!5^n}{2(n+1)!5^n}$

    Simplified

    $\displaystyle tn+1/tn = \frac{2n+24*5^n}{2n+2*5^n}$


    I am a bit lost from this point. Can you please tell me if I am even on the right track?
    Last edited by Newbie999; Mar 21st 2013 at 11:21 PM.
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  4. #4
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    Re: Sequence with factorial

    Quote Originally Posted by Newbie999 View Post
    $\displaystyle tn+1/tn = \frac{2(n+2)!/5^n}{2(n+1)!5^n}$
    is incorrect, but close. you also needed to replace n with n+1 in the exponent of 5 on the numerator

    $\displaystyle \frac{t_{n+1}}{t_n} = \frac{2(n+2)!5^{n+1}}{2(n+1)!5^n}$

    Notice that the 2's cancel out. and that
    $\displaystyle \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)!}{(n+1)!}} = (n+2)$ by definition of factorial
    $\displaystyle \frac{5^{n+1}}{5^n} = \frac{5*5^n}{5^n} = 5$

    hence

    $\displaystyle \frac{t_{n+1}}{t_n} = \frac{2(n+2)!5^{n+1}}{2(n+1)!5^n} = 5(n+2)$
    Thanks from Newbie999 and smokesalot
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  5. #5
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    Re: Sequence with factorial

    Thank you sooo much : ) for everything.

    Edit: One more question. I would like to send you a PM since this is solved but someone might found it useful to see the question here.


    With the $\displaystyle tn+1/tn = \frac{5{^{n+1}}}{5^{n}}$
    I pretty much understand everything above and the final answer.

    But what if $\displaystyle tn-1/tn = \frac{5{^{n-1}}}{5^{n}}$

    How can I calculate a sequence when it's relating to previous number? Lets say n-1 or n-2.
    Last edited by Newbie999; Mar 21st 2013 at 11:39 PM.
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  6. #6
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    Re: Sequence with factorial

    If the general formula for $\displaystyle t_n$ was $\displaystyle 5^n$ instead, then

    $\displaystyle \frac{t_{n-1}}{t_n} = \frac{5^{n-1}}{5^n} = \frac{1}{5}$ hence $\displaystyle t_{n-1} = \frac{1}{5}t_n$
    so this is still a relationship between two consecutive numbers. Replace n with n-1 and you get a relationship between $\displaystyle t_{n-1}$ and $\displaystyle t_{n-2}$
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  7. #7
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    Re: Sequence with factorial

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