Math Help - Binomial Coefficient

1. Originally Posted by PaulRS
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 $
\sum\limits_{k = 0}^n {\delta \left( k \right)} = \left\lfloor {\tfrac{n}
{5}} \right\rfloor + 1
$
where $
\left\lfloor x \right\rfloor
$
is the floor function

So the coefficient of $x^{100}$ turns out to be $
\sum\limits_{k = 0}^{100} {\delta \left( k \right)} = \left\lfloor {\tfrac{{100}}
{5}} \right\rfloor + 1 = 21
$
whered you get the +1?

2. Originally Posted by p00ndawg
whered you get the +1?
0 is multiple of 5

Page 2 of 2 First 12