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Math Help - Binomial Coefficient

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

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

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

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

    So the coefficient of x^{100} turns out to be <br />
\sum\limits_{k = 0}^{100} {\delta \left( k \right)}  = \left\lfloor {\tfrac{{100}}<br />
{5}} \right\rfloor  + 1 = 21<br />
    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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