# Binomial Coefficient

#### p00ndawg

$\displaystyle (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})$

I need the coefficient of $\displaystyle x^{100}$

can anyone help?

If you have any tips for solving these kinds of problems, it would be really appreciated.

#### Mathstud28

I Think

$\displaystyle (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})$

I need the coefficient of $\displaystyle x^{100}$

can anyone help?

If you have any tips for solving these kinds of problems, it would be really appreciated.
The answer is N/A...there is no $\displaystyle x^100$ term after you expand

#### p00ndawg

The answer is N/A...there is no $\displaystyle x^100$ term after you expand
wat?

it should be 21. There has to be a term, $\displaystyle (1/1-x) = (1+ x+ x^{2}+x^{3}+x^{4}+...)$

#### Mathstud28

O....

wat?

it should be 21. There has to be a term, $\displaystyle (1/1-x) = (1++ x+ x^{2}+x^{3}+x^{4}+...)$
Do you mean the power series for $\displaystyle \frac{1}{1-x}$...you know that only applys for $\displaystyle -1<x<1$ right? you cant make a general statement unless you also state that?

#### p00ndawg

Do you mean the power series for $\displaystyle \frac{1}{1-x}$...you know that only applys for $\displaystyle -1<x<1$ right? you cant make a general statement unless you also state that?
well originally the question was find the number of ways to make change for 1 dollar using up to 9 pennies and any number of nickels and dimes.

I just figured it was implied, my bad.

I just reduced it about as far as I can, but I dont know how to get the number of coefficients easily.

not sure if theres an easy way to do it, but my professor did it in his head lol.

#### Mathstud28

Hmm

well originally the question was find the number of ways to make change for 1 dollar using up to 9 pennies and any number of nickels and dimes.

I just figured it was implied, my bad.

I just reduced it about as far as I can, but I dont know how to get the number of coefficients easily.

not sure if theres an easy way to do it, but my professor did it in his head lol.
That sounds more like a $\displaystyle _nC_r$ problem...yeah?

#### p00ndawg

That sounds more like a $\displaystyle _nC_r$ problem...yeah?
its a generating function problem, cleverly disguised as a binomial coefficient problem, cleverly disguised as an innocent, yet disturbing way of dividing up a dollar.

#### Mathstud28

its a generating function problem, cleverly disguised as a binomial coefficient problem, cleverly disguised as an innocent, yet disturbing way of dividing up a dollar.
are you sure that it is the power series for $\displaystyle \frac{1}{1-x}$?

#### p00ndawg

are you sure that it is the power series for $\displaystyle \frac{1}{1-x}$?
yea, I originally had the problem like this..

$\displaystyle (1 + x + x^{2} +...+x^{9})(1+x^{5}+x^{10}+x^{15}+...+x^{100})(1+x^{10}+x^{20}+x^{30}+...+x^{100})$

I then reduced the first expression to :
$\displaystyle (1-x^{10}/1-x)$ left the second expression untouched, and reduced the third to, $\displaystyle (1/1-x^{10})$ .

I actually have the problem worked out, but the last line where the professor got the answer left me blank, because I cant see how he got 21 so easily.

#### Mathstud28

haha

yea, I originally had the problem like this..

$\displaystyle (1 + x + x^{2} +...+x^{9})(1+x^{5}+x^{10}+x^{15}+...+x^{100})(1+x^{10}+x^{20}+x^{30}+...+x^{100})$

I then reduced the first expression to :
$\displaystyle (1-x^{10}/1-x)$ left the second expression untouched, and reduced the third to, $\displaystyle (1/1-x^{10})$ .

I actually have the problem worked out, but the last line where the professor got the answer left me blank, because I cant see how he got 21 so easily.
Ok this may be a really stupid question...but did he show you some step that gave the right answer or did he just know the answer? if it is the latter did you ever consider this is a planned out problem?