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Thread: Binomial Coefficient

  1. #16
    Member
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    Jan 2008
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    Quote Originally Posted by PaulRS View Post
    Let: $\displaystyle \delta(n)=0$ if 5 doesn't divide n, and $\displaystyle \delta(n)=1$ if 5 does divide n

    The coefficient of $\displaystyle x^{100}$ stays untouched if we change $\displaystyle (1+x^{5}+x^{10}+x^{15}+...+x^{100})$ for $\displaystyle 1+x^{5}+x^{10}+x^{15}+...$

    We have $\displaystyle 1+x^{5}+x^{10}+x^{15}+...=\sum_{k=0}^{\infty}{\del ta(n)\cdot{x^n}}$

    Thus: $\displaystyle (1+x+x^{2}+...)(1+x^{5}+x^{10}+x^{15}+...)=\sum_{k =0}^{\infty}{\sum_{k=0}^n{\delta(k)}}\cdot{x^n}$

    But: $\displaystyle \sum_{k=0}^n{\delta(k)}$ is the number of multiples of 5 between 0 and n, thus we have $\displaystyle
    \sum\limits_{k = 0}^n {\delta \left( k \right)} = \left\lfloor {\tfrac{n}
    {5}} \right\rfloor + 1
    $ where $\displaystyle
    \left\lfloor x \right\rfloor
    $ is the floor function

    So the coefficient of $\displaystyle x^{100}$ turns out to be $\displaystyle
    \sum\limits_{k = 0}^{100} {\delta \left( k \right)} = \left\lfloor {\tfrac{{100}}
    {5}} \right\rfloor + 1 = 21
    $
    whered you get the +1?
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  2. #17
    Super Member PaulRS's Avatar
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    Quote Originally Posted by p00ndawg View Post
    whered you get the +1?
    0 is multiple of 5
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