$\displaystyle (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})$
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.
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.
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.
He explained it, and told us the logic to what the answer was, undeniably I couldnt really write down what he was saying, but what he said made sense.
I just cant figure it out.
It was also a review problem for my test tomorrow but these binomial coefficient problems are kicking my ass.
Can you help me with this problem then..
$\displaystyle (x^{3}+x^{4}+x^{5}+x^{6}+x^{7}...)^{3}$
I need the coefficient of $\displaystyle x^{10}$
Yep, it's 3. You always have to choose one term from 1st product, one from the second and one from the 3rd, but in order to get $\displaystyle x^{10}$, two of these must be $\displaystyle x^{3}$ and the other $\displaystyle x^4$. Thus the number of permutations of $\displaystyle (x^3)(x^3)(x^4)$ is your coefficient.
Let: $\displaystyle \delta(n)=0$ if 5 doesn't divide n, and $\displaystyle \delta(k)=1$ 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 $\displaystyle 1+x^{5}+x^{10}+x^{15}+...=\sum_{n=0}^{\infty}{\del ta(n)\cdot{x^n}}$
Thus: $\displaystyle (1+x+x^{2}+...)(1+x^{5}+x^{10}+x^{15}+...)=\sum_{n =0}^{\infty}{\sum_{k=0}^n{\delta(k)}}\cdot{x^n}$
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
$
Stated in another way: For each $\displaystyle 1,x^{5},...,x^{100}$, you have only one element among $\displaystyle 1,x^{1},x^{2},...$ such that the product of the pair gives$\displaystyle x^{100}$. It follows then that the coefficient of $\displaystyle x^{100}$ is actually 21, which is the numbers of terms of the factor $\displaystyle (1+x^5+...+x^{100})$