find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$  

The idea is to use the formula
$$
\frac{1}{(1-x)^t} = \sum_{n=0}^\infty \binom{n+t-1}{n} x^n,
$$
which can be proved by induction on $t$.
Using this we get
$$
\frac{1+x^4}{(1+2x)^3(1-2x)^2} = \\
(1+O(x^4))(1 - 6x + 24x^2 - 80x^3+O(x^4))(1 + 4x + 12x^2 + 32x^3 + O(x^4)) = \\
1 + (-6+4)x + (24-6\cdot 4+12)x^2 + (-80 + 24\cdot 4 - 6\cdot 12 + 32)x^3 +
O(x^4) = \\
1 - 2x + 12x^2 - 24x^3 + O(x^4).
$$
