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Math Help - challenging factorial problem

  1. #1
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    challenging factorial problem

    Simplify

    2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!,

    where k! denotes the product of all integers from 1 up to k,

    that is: 1! = 1, 2! =1 * 2 = 2, 3! = 3*2*1 = 6 and so on...
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  2. #2
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    Tn = ((n+1)/n!)(-1)^{n+1}<br />
=(-1)^{n+1}(1/(n-1)! + 1/n!)<br />

    n varies from 1 to 2010, therefore, the series reduces to,
    1+1 -1 -1/2!+1/2!+1/3!-1/3!-1/4!+............-1/2010!

    All the intermediate terms vanish, and what survives is
    S=1-1/2010!

    Thus the value of this expression is 1-1/2010!
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  3. #3
    Super Member
    Joined
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    you might want to try the expansion of

    xe^x

    Quote Originally Posted by the undertaker View Post
    Simplify

    2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!,

    where k! denotes the product of all integers from 1 up to k,

    that is: 1! = 1, 2! =1 * 2 = 2, 3! = 3*2*1 = 6 and so on...
    Follow Math Help Forum on Facebook and Google+

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