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Math Help - 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

    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
    Senior Member MacstersUndead's Avatar
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    Re: Sequence with factorial

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

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

    Since we know the formula for t_n we also know the formula for t_{n+1} (replace n with n+1)
    Now consider \frac{t_{n+1}}{t_n}, the ratio between two consecutive terms, and isolate 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.

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

    Simplified

    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; March 21st 2013 at 11:21 PM.
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    Senior Member MacstersUndead's Avatar
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    Re: Sequence with factorial

    Quote Originally Posted by Newbie999 View Post
    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

    \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
    \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)!}{(n+1)!}} = (n+2) by definition of factorial
    \frac{5^{n+1}}{5^n} = \frac{5*5^n}{5^n} = 5

    hence

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

    But what if 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; March 21st 2013 at 11:39 PM.
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  6. #6
    Senior Member MacstersUndead's Avatar
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    Re: Sequence with factorial

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

    \frac{t_{n-1}}{t_n} = \frac{5^{n-1}}{5^n} = \frac{1}{5} hence 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 t_{n-1} and t_{n-2}
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    Re: Sequence with factorial

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