What is the coefficient of $x^{11}$ in the power series expansion of $\frac{1}{1-x-x^4}$? I am really stuck on this problem. I don't really understand power series expansions. However, I think this has to do with generating functions.
 A: $\frac {1}{1-x} = \sum x^n\\
\frac {1}{1-(x+x^4)} = \sum (x+x^4)^n$
$x^{11}$ appears in the expansion of  $(x+x^4)^5, (x+x^4)^8,(x+x^4)^{11}$
${5\choose 2}+{8\choose 1} + {11\choose 0} = 19$
A: $\frac{1}{1-x-x^4}=\frac{1}{1-(x+x^4)}=1+(x+x^4)+(x+x^4)^2+(x+x^4)^3+(x+x^4)^4+(x+x^4)^5+(x+x^4)^6+(x+x^4)^7+(x+x^4)^8+(x+x^4)^9+(x+x^4)^{10}+(x+x^4)^{11}+\cdot\cdot\cdot$.
Checking how many $x^{11}$s are contributed by each term and adding up, we see the answer is $0+0+0+0+0+\binom{5}{2}+0+0+\binom{8}{1}+0+0+\binom{11}{0}=10+8+1$.
So the answer is 19.
A: The power serie associated to $\frac{1}{1-u}$ is $\sum_{k\geq 0}u^k$. Using this formula with $u=x+x^4$, we get $$\frac{1}{1-x-x^4}=\sum_{k\geq 0}(x+x^4)^k$$ 
We want to gather every term $x^{11}$ in the developping of this sum. First, let us write down $$(x+x^4)^k=\sum_{l=0}^{k}\dbinom{k}{l}x^{3l + k}$$ by Newton's formula. Checking by hand, we see that a term $x^{11}$ only occurs in the cases $(k,l)\in \{(11,0);(8,1);(5,2)\}$, whith associated coefficients $\dbinom{11}{0}=1, \dbinom{8}{1}=8, \dbinom{5}{2}=10$.  
Summing them, we conclude that the coefficient in front of $x^{11}$ is $19$.
