n次方差公式：
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a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+···+ab^{n-2}+b^{n-1})，n \in N^{*}
an−bn=(a−b)(an−1+an−2b+an−3b2+⋅⋅⋅+abn−2+bn−1)，n∈N∗
证法一：
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\begin{aligned} a^{n}-b^{n} &= a^{n}-a^{n-1}b+a^{n-1}b-a^{n-2}b^{2}+a^{n-2}b^{2}-···+ab^{n-1}-b^{n} \\ &= a^{n-1}(a-b)+a^{n-2}b(a-b)+···+b^{n-1}(a-b) \\ &= (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+···+ab^{n-2}+b^{n-1}) \end{aligned}
an−bn=an−an−1b+an−1b−an−2b2+an−2b2−⋅⋅⋅+abn−1−bn=an−1(a−b)+an−2b(a−b)+⋅⋅⋅+bn−1(a−b)=(a−b)(an−1+an−2b+an−3b2+⋅⋅⋅+abn−2+bn−1)
证法二：
设等比数列
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an 的通项公式为
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a_{n}=(\frac{b}{a})^{n}
an=(ab)n ，则其前
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\begin{aligned} & \frac{b}{a}+(\frac{b}{a})^{2}+(\frac{b}{a})^{3}+···+(\frac{b}{a})^{n-1}+(\frac{b}{a})^{n} \\ & = \frac{\frac{b}{a}[1-(\frac{b}{a})^{n}]}{1-\frac{b}{a}} = \frac{b[1-(\frac{b}{a})^{n}]}{a-b} = \frac{b(a^{n}-b^{n})}{a^{n}(a-b)} \end{aligned}
ab+(ab)2+(ab)3+⋅⋅⋅+(ab)n−1+(ab)n=1−abab[1−(ab)n]=a−bb[1−(ab)n]=an(a−b)b(an−bn)
故：
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\begin{aligned} a^{n}-b^{n} &= \frac{a^{n}(a-b)}{b}[\frac{b}{a}+(\frac{b}{a})^{2}+(\frac{b}{a})^{3}+···+(\frac{b}{a})^{n-1}+(\frac{b}{a})^{n}] \\ &= (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+···+ab^{n-2}+b^{n-1}) \end{aligned}
an−bn=ban(a−b)[ab+(ab)2+(ab)3+⋅⋅⋅+(ab)n−1+(ab)n]=(a−b)(an−1+an−2b+an−3b2+⋅⋅⋅+abn−2+bn−1)文章来源地址https://www.toymoban.com/news/detail-532046.html
文章来源：https://www.toymoban.com/news/detail-532046.html
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