7. Find the coefficient of $x^{12}$ in the Maclaurin series (i.e. Taylor series around $x=0$ ) for
$$
\frac{1}{1-3 x+2 x^{2}} \text {. }
$$
Solution: If you know formal power series, then this is not such a hard question, but since this is a calculus test... Use partial fractions to get $\frac{1}{1-3 x+2 x^{2}}=\frac{1 / 2}{1-2 x}-\frac{1}{1-x}$. Now each of these can be expanded as a geometric series (or take derivatives and get the same result) to get $\frac{1}{2}\left(1+2 x+4 x^{2}+8 x^{3}+\cdots\right)-\left(1+x+x^{2}+x^{3}+\cdots\right)$, so the coefficient of $x^{n}$ is $2^{n-1}-1$. When $n=12$, that's 2047 .