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n次方差公式：
a n − b n = ( a − b ) ( a n − 1 + a n − 2 b + a n − 3 b 2 + ⋅ ⋅ ⋅ + a b n − 2 + b n − 1 ) ， n ∈ N ∗ 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：
a n − b n = a n − a n − 1 b + a n − 1 b − a n − 2 b 2 + a n − 2 b 2 − ⋅ ⋅ ⋅ + a b 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 + ⋅ ⋅ ⋅ + a b n − 2 + b n − 1 ) \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 a n {a_{n}} an 的通项公式为 a n = ( b a ) n a_{n}=(\frac{b}{a})^{n} an=(ab)n ，则其前 n n n 项和为：
b a + ( b a ) 2 + ( b a ) 3 + ⋅ ⋅ ⋅ + ( b a ) n − 1 + ( b a ) n = b a [ 1 − ( b a ) n ] 1 − b a = b [ 1 − ( b a ) n ] a − b = b ( a n − b n ) a n ( a − b ) \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：
a n − b n = a n ( a − b ) b [ b a + ( b a ) 2 + ( b a ) 3 + ⋅ ⋅ ⋅ + ( b a ) n − 1 + ( b a ) n ] = ( a − b ) ( a n − 1 + a n − 2 b + a n − 3 b 2 + ⋅ ⋅ ⋅ + a b n − 2 + b n − 1 ) \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)
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