记住核心的展开式然后通过变形求解
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\frac{1}{1-x} = \sum_{n=0}^{\infty } x^{n} "><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\frac{1}{1-x^2} = \sum_{n=0}^{\infty } x^{2n} "><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\frac{1}{1+x} = \sum_{n=0}^{\infty } (-1)^nx^{n} "><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\frac{1}{1+x^2} = \sum_{n=0}^{\infty } (-1)^nx^{2n}"><span></span><span></span></span>
记忆技巧，都是基于第一条来记忆，如果符号不变次幂变了，则用n次幂×x的次幂
如果符号变＋号，则前面新增（-1）的n次幂，后面x的次幂参考上面
一定要记住x范围是-1<x<1
e的次幂展开式
<span class="equation-text" contenteditable="false" data-index="0" data-equation="e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!} （-\infty <x<+\infty）"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="e^{2x} = \sum_{n=0}^{\infty } \frac{2x^n}{n!} （-\infty <x<+\infty）"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="e^{x^2} = \sum_{n=0}^{\infty } \frac{x^{2n}}{n!} （-\infty <x<+\infty）"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="e^{-x^2} = \sum_{n=0}^{\infty } \frac{(-1)^nx^{2n}}{n!} （-\infty <x<+\infty）"><span></span><span></span></span>
记忆技巧，x直接替换掉,x就是x的n次幂,2x就是2x的n次幂，如此类推，如果有10x那就是10的n次幂
