问题及解答

(1) $\frac{1}{1+x^2}$.

(2) $\frac{1}{x^2-5x+6}$.

(1)

\[
\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-x^2)^n=1-x^2+x^4-x^6+x^8-\cdots,\quad x\in(-1,1).
\]

(2)

\[
\begin{split}
\frac{1}{x^2-5x+6}&=\frac{1}{(x-2)(x-3)}=\frac{1}{(2-x)}\cdot\frac{1}{(3-x)}\\
&=\frac{1}{2-x}-\frac{1}{3-x}=\frac{1}{2}\cdot\frac{1}{1-\frac{x}{2}}-\frac{1}{3}\cdot\frac{1}{1-\frac{x}{3}}\\
&=\frac{1}{2}\sum_{n=0}^{\infty}(\frac{x}{2})^n-\frac{1}{3}\sum_{n=0}^{\infty}(\frac{x}{3})^n\\
&=\sum_{n=0}^{\infty}(\frac{1}{2^{n+1}}-\frac{1}{3^{n+1}})x^n,
\end{split}
\]

这里 $x$ 需满足 $|\frac{x}{2}| < 1$ 且 $|\frac{x}{3}| < 1$, 从而 $x$ 的范围是 $(-2,2)$.