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等幂和差 Connected to: {{::readMoreArticle.title}} 等幂和差,又称幂和差,指同是n次幂的a与b的和与差,即 a n ± b n {\displaystyle a^{n}\pm b^{n)) 。 等比数列求和公式的推导[编辑]等幂和[编辑]当n为奇数时有: a n + b n = ( a + b ) ∑ r = 1 n a n − r ( − b ) r − 1 = ( a + b ) ( a n − 1 − a n − 2 b + . . . + b n − 1 ) {\displaystyle a^{n}+b^{n}=(a+b)\sum _{r=1}^{n}a^{n-r}(-b)^{r-1}=(a+b)(a^{n-1}-a^{n-2}b+...+b^{n-1})} 因尾项必须为+bn-1,故n不能取偶数。 等幂差[编辑]首项为1、公比为q的等比数列求和后得出: ∑ r = 1 n q r − 1 = 1 + q + . . . + q n − 1 = q n − 1 q − 1 {\displaystyle \sum _{r=1}^{n}q^{r-1}=1+q+...+q^{n-1}={\frac {q^{n}-1}{q-1))} 设q=a/b,换算后即有: a n − b n = ( a − b ) ∑ r = 1 n a n − r b r − 1 = ( a − b ) ( a n − 1 + a n − 2 b + . . . + b n − 1 ) {\displaystyle a^{n}-b^{n}=(a-b)\sum _{r=1}^{n}a^{n-r}b^{r-1}=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})} 逆定理[编辑] ∑ r = 1 n a n − r b r − 1 = a n − 1 + a n − 2 b + . . . + b n − 1 = a n − b n a − b {\displaystyle \sum _{r=1}^{n}a^{n-r}b^{r-1}=a^{n-1}+a^{n-2}b+...+b^{n-1}={\frac {a^{n}-b^{n)){a-b))} ∑ r = 1 n a n − r ( − b ) r − 1 = a n − 1 − a n − 2 b + . . . + b n − 1 = a n + b n a + b {\displaystyle \sum _{r=1}^{n}a^{n-r}(-b)^{r-1}=a^{n-1}-a^{n-2}b+...+b^{n-1}={\frac {a^{n}+b^{n)){a+b))} 注意 ∑ r = 1 2 n − 1 a 4 n − 2 − 2 r b 2 r − 2 {\displaystyle \sum _{r=1}^{2n-1}a^{4n-2-2r}b^{2r-2)) 可进行因式分解,例如: a 4 + a 2 b 2 + b 4 = a 6 − b 6 a 2 − b 2 = ( a 3 + b 3 ) ( a 3 − b 3 ) ( a + b ) ( a − b ) = ( a 2 + a b + b 2 ) ( a 2 − a b + b 2 ) {\displaystyle a^{4}+a^{2}b^{2}+b^{4}={\frac {a^{6}-b^{6)){a^{2}-b^{2))}={\frac {(a^{3}+b^{3})(a^{3}-b^{3})}{(a+b)(a-b)))=(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})} 另解: a 4 + a 2 b 2 + b 4 = a 4 + 2 a 2 b 2 + b 4 − a 2 b 2 = ( a 2 + b 2 ) 2 − a 2 b 2 = ( a 2 + a b + b 2 ) ( a 2 − a b + b 2 ) {\displaystyle a^{4}+a^{2}b^{2}+b^{4}=a^{4}+2a^{2}b^{2}+b^{4}-a^{2}b^{2}=(a^{2}+b^{2})^{2}-a^{2}b^{2}=(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})} 参见[编辑]卢卡斯数列 等幂求和查论编基本乘法公式及恒等式 (因式分解)分配律 ( a + b ) ( c + d ) = a c + a d + b c + b d {\displaystyle (a+b)(c+d)=ac+ad+bc+bd\,\!} 二项式定理和与差的平方 ( a ± b ) 2 = a 2 ± 2 a b + b 2 {\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2)) 和与差的立方 ( a ± b ) 3 = a 3 ± 3 a 2 b + 3 a b 2 ± b 3 {\displaystyle (a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3)) 多项式定理三数和平方 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a {\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\,\!} 等幂和差平方和 a 2 + b 2 = ( a + b i ) ( a − b i ) {\displaystyle a^{2}+b^{2}=(a+bi)(a-bi)} 平方差 a 2 − b 2 = ( a + b ) ( a − b ) {\displaystyle a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)} 立方和和立方差 a 3 ± b 3 = ( a ± b ) ( a 2 ∓ a b + b 2 ) {\displaystyle a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})\,\!} 等幂和差逆定理 a 4 + a 2 b 2 + b 4 = ( a 2 + a b + b 2 ) ( a 2 − a b + b 2 ) {\displaystyle a^{4}+a^{2}b^{2}+b^{4}=(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})\,\!} 对称多项式 a 3 + b 3 + c 3 = ( a + b + c ) 3 + 3 ( a + b + c ) ( − a b − b c − c a ) + 3 a b c {\displaystyle a^{3}+b^{3}+c^{3}=(a+b+c)^{3}+3(a+b+c)(-ab-bc-ca)+3abc\,\!} 分类 分类:数学公式初等代数 {{bottomLinkPreText}} {{bottomLinkText}} This page is based on a Wikipedia article written by contributors (read/edit). Text is available under the CC BY-SA 4.0 license; additional terms may apply. Images, videos and audio are available under their respective licenses. Cover photo is available under {{::mainImage.info.license.name || 'Unknown'}} license. Cover photo is available under {{::mainImage.info.license.name || 'Unknown'}} license. Credit: (see original file).