I've been stuck on this question for a little while, could someone point me in the right direction? I'm supposed to determine the coefficient of $x^8$.
$$x^8\quad in \quad \frac{x}{(1-x)(1-2x)} $$
I know I have to use binomial theorem however, I'm finding it difficult to get started.
You can use the partial fraction technic to split it into two: $$ \dfrac{x}{(1-x)(1-2x)} = \dfrac{A}{1-x} + \dfrac{B}{1-2x} $$
Which yields the following equation: $$ A (1-2x) + B(1-x) = x $$ Subbing $x=1$ we get $A=-1$, and subbing in $x=\dfrac{1}{2}$ we get $B = 1$ and thus we have $$ \dfrac{x}{(1-x)(1-2x)} = \dfrac{1}{1-2x} - \dfrac{1}{1-x} $$ Now we know that $\dfrac{1}{1-2x} = \sum \limits_{n=0}^{\infty} (2x)^n$, and $\dfrac{1}{1-x} = \sum \limits_{n=0}^{\infty} x^n$ (the closed form of generating functions), hence we have now $$ \dfrac{x}{(1-x)(1-2x)} = \sum \limits_{n=0}^{\infty} (2x)^n - \sum \limits_{n=0}^{\infty} x^n $$
Now finding the coefficient of $x^8$ is simple, plug in $n=8$ and we get the coefficient for $x^8$ is $2^8 - 1$ which is 255. You don't really need the binomial theorem here, just a brush up on your generating functions tricks.
Cheers
you might find it easier if you express it in partial fractions first, i.e. $$\frac {-1}{(1-x)}+\frac {1}{(1-2x)}$$
Now it's going to be a lot easier to find the term you need
Write as $$x(1-x)^{-1}(1-2x)^{-1}$$
Usaing the binomial theorem on $(1-x)^{-1}=(1+(-x))^{-1}$ gives \begin{eqnarray} (1+(-x))^{-1} &=& 1-(-x)+(-x)^{2}-(-x)^{3}+(-x)^{4}\ldots \end{eqnarray} Then do the same for $(1-2x)^{-1}$ and when you come to multiply out the results you just pick the ways in which you can form a $x^{8}$ coefficient. There aren't many cases to consider.
Use the http://en.wikipedia.org/wiki/Cauchy_product for series: $$ \frac{x}{(1-x)(1-2x)} = x \sum_{k=0}^{\infty}x^k\sum_{k=0}^{\infty}(2x)^k$$ $$=x\sum_{j=0}^{\infty} c_j x^j\quad\text{with}\quad c_j = \sum_{k=0}^{j}a_k b_{j-k}$$ where $a_k=1, b_k=2^k$. The factor $x$ shifts the indices, therefore the coefficient for $x^s$ for $s>0$ is $$\sum_{j=0}^{s-1}2^j=2^{s}-1$$
