EM@等幂和差公式
文章目录
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- abstract
- 等幂差
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- 等比数列性质推导
- 二项式展开的思路
- 等幂和
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- 奇次方幂之和
- 偶次方幂之和
- 证明等幂差
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abstract
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讨论整数指数幂中等幂和差的展开
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这里讨论的等幂指的是指数相等的两个幂
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并且仅在整数指数幂范围内讨论,实指数幂范围内此处不讨论
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两个n次方幂值差展开公式
- a n − b n a^n-b^n an−bn= ( a − b ) ( a n − 1 b 0 + a n − 2 b 1 + ⋯ + a 1 b n − 2 + a 0 b n − 1 ) (a-b)(a^{n-1}b^{0}+a^{n-2}b^{1}+\cdots+a^{1}b^{n-2}+a^{0}b^{n-1}) (a−b)(an−1b0+an−2b1+⋯+a1bn−2+a0bn−1)
- 两个立方数值差 a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3-b^3=(a-b)(a^2+ab+b^2) a3−b3=(a−b)(a2+ab+b2)
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两个奇数次方数之和展开公式
- a n + b n a^n+b^n an+bn= ( a + b ) ( a n − 1 b 0 − a n − 2 b 1 + ⋯ + a 0 b n − 1 ) (a+b)(a^{n-1}b^{0}-a^{n-2}b^{1}+\cdots+a^0b^{n-1}) (a+b)(an−1b0−an−2b1+⋯+a0bn−1)
- 两个立方数之和 a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3=(a+b)(a^2-ab+b^2) a3+b3=(a+b)(a2−ab+b2)
- 其中等幂和可以由等幂差公式直接变式得到
等幂差
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等幂差展开公式 a n − b n a^{n}-b^{n} an−bn可以展开为如下公式
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a n − b n a^{n}-b^{n} an−bn ( n ∈ N + ) {(n\in\mathbb{N^+})} (n∈N+)
- = ( a − b ) ( a n − 1 b 0 + a n − 2 b 1 + ⋯ + a 1 b n − 2 + a 0 b n − 1 ) (a-b)(a^{n-1}b^{0}+a^{n-2}b^{1}+\cdots+a^{1}b^{n-2}+a^{0}b^{n-1}) (a−b)(an−1b0+an−2b1+⋯+a1bn−2+a0bn−1)
- = ( a − b ) ∑ i = 0 n − 1 a n − 1 − i b i (a-b) \sum_{i=0}^{n-1} a^{n-1-i}b^{i} (a−b)∑i=0n−1an−1−ibi
- = ( a − b ) ∑ i = 1 n a n − i b i − 1 (a-b) \sum_{i=1}^{n} a^{n-i}b^{i-1} (a−b)∑i=1nan−ibi−1
- = ( a − b ) ∑ i = 0 θ a θ − i b i (a-b)\sum\limits_{i=0}^{\theta}a^{\theta-i}b^{i} (a−b)i=0∑θaθ−ibi, ( θ = n − 1 ) , ( θ ∈ { 2 , 3 , ⋯ } ) {(\theta=n-1)},{(\theta\in{\{2,3,\cdots\}})} (θ=n−1),(θ∈{2,3,⋯})
- = ( a − b ) ∑ r 1 + r 2 = n − 1 a r 1 b r 2 (a-b)\sum\limits_{r_1+r_2=n-1}a^{r_1}b^{r_2} (a−b)r1+r2=n−1∑ar1br2, ( r 1 , r 2 ∈ Z ) {(r_1,r_2\in\mathbb{Z})} (r1,r2∈Z)
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例如 n = 3 n=3 n=3时: a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3-b^3=(a-b)(a^2+ab+b^2) a3−b3=(a−b)(a2+ab+b2)
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公式特点
- 公式右边分为2部分, a − b a-b a−b和求和部分,求和式共有 n n n项,每项的幂都是 n − 1 n-1 n−1次幂
- 上述5种书写形式以 ( 5 ) (5) (5)最为优雅;推理的时候形式 ( 3 ) (3) (3)最为协调,
- 形似(1)不使用连加号而使用省略号表示可能使最有利于发现规律和简化的
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可以结合多项式的次数和余式定理来理解记忆公式
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应用:整数幂不等式性质: a > b > 0 ⇒ a n > b n a>b>0\Rightarrow{a^n>b^n} a>b>0⇒an>bn, n ∈ N + , n ⩾ 2 n\in\mathbb{N^+},n\geqslant2 n∈N+,n⩾2
等比数列性质推导
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对于 a n − b n a^{n}-b^{n} an−bn,构造等比数列 { a n } \set{a_n} {an}, a 1 = 1 a_1=1 a1=1, q = a b q=\frac{a}{b} q=ba;令数列 { a n } \set{a_n} {an}的前 n n n项和为 S n S_n Sn,
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当首项 a 1 = 1 a_1=1 a1=1的时候,等比数列前 n n n项和退化为更加简单的形式:
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S n S_n Sn= ∑ i = 1 n 1 ⋅ q i − 1 \sum\limits_{i=1}^{n}1\cdot q^{i-1} i=1∑n1⋅qi−1= q 0 + q 1 + q 2 + ⋯ + q n − 1 q^0+q^1+q^2+\dots+q^{n-1} q0+q1+q2+⋯+qn−1= q n − 1 q − 1 \frac{q^n-1}{q-1} q−1qn−1
(1) -
将 S n S_n Sn表示为 a , b a,b a,b的求和式 S n = ∑ i = 0 n − 1 q i S_n=\sum\limits_{i=0}^{n-1}q^i Sn=i=0∑n−1qi= ∑ i = 0 n − 1 a i b i \sum\limits_{i=0}^{n-1}\frac{a^i}{b^i} i=0∑n−1biai= ∑ i = 0 n − 1 a i b − i \sum\limits_{i=0}^{n-1}{a^i}{b^{-i}} i=0∑n−1aib−i
(2)
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(1)代入 q = a b q=\frac{a}{b} q=ba,得到 S n S_n Sn= q n − 1 q − 1 \frac{q^n-1}{q-1} q−1qn−1= a n b n − 1 a b − 1 \Large\frac{\frac{a^n}{b^n}-1}{\frac{a}{b}-1} ba−1bnan−1= a n − b n b n a − b b \Large\frac{\frac{a^n-b^n}{b^n}}{\frac{a-b}{b}} ba−bbnan−bn= a n − b n b n − 1 ( a − b ) \frac{a^n-b^n}{b^{n-1}(a-b)} bn−1(a−b)an−bn, -
从而 a n − b n a^{n}-b^{n} an−bn= S n b n − 1 ( a − b ) S_{n}b^{n-1}(a-b) Snbn−1(a−b)
- b n − 1 S n b^{n-1}S_n bn−1Sn= b n − 1 ∑ i = 0 n − 1 a i b − i b^{n-1}\sum\limits_{i=0}^{n-1}{a^i}{b^{-i}} bn−1i=0∑n−1aib−i= ∑ i = 0 n − 1 a i b − i b n − 1 \sum\limits_{i=0}^{n-1}{a^i}{b^{-i}}b^{n-1} i=0∑n−1aib−ibn−1= ∑ i = 0 n − 1 a i b n − 1 − i \sum\limits_{i=0}^{n-1}{a^i}{b^{n-1-i}} i=0∑n−1aibn−1−i
- a n − b n = b n − 1 S n ( a − b ) a^n-b^n=b^{n-1}S_n(a-b) an−bn=bn−1Sn(a−b)= ( a − b ) ∑ i = 0 n − 1 a i b n − 1 − i (a-b)\sum\limits_{i=0}^{n-1}{a^i}{b^{n-1-i}} (a−b)i=0∑n−1aibn−1−i= ( a − b ) ∑ i = 0 n − 1 a n − 1 − i b i (a-b)\sum\limits_{i=0}^{n-1}{a^{n-1-i}}{b^i} (a−b)i=0∑n−1an−1−ibi
- = ( a − b ) ∑ i = 1 i = n a n − i b i − 1 (a-b)\sum\limits_{i=1}^{i=n}a^{n-i}b^{i-1} (a−b)i=1∑i=nan−ibi−1
- = ( a − b ) ∑ r 1 + r 2 = n − 1 a r 1 b r 2 (a-b)\sum\limits_{r_1+r_2=n-1}a^{r_1}b^{r_2} (a−b)r1+r2=n−1∑ar1br2, ( r 1 , r 2 ∈ Z ) {(r_1,r_2\in\mathbb{Z})} (r1,r2∈Z)
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特别的, n = 3 n=3 n=3时, a 3 − b 3 a^3-b^3 a3−b3= ( a − b ) ∑ i = 1 3 a 3 − i b i − 1 (a-b)\sum\limits_{i=1}^{3}a^{3-i}b^{i-1} (a−b)i=1∑3a3−ibi−1= ( a − b ) ( a 2 b 0 + a 1 b 1 + a 0 b 2 ) (a-b)(a^2b^0+a^1b^1+a^0b^2) (a−b)(a2b0+a1b1+a0b2)
二项式展开的思路
- 令 c = a − b c=a-b c=a−b,则 a = b + c a=b+c a=b+c
- 从而 a n − b n a^{n}-b^{n} an−bn= ( b + c ) n − b n (b+c)^n-b^{n} (b+c)n−bn= ∑ k = 0 n ( n k ) b k c n − k − b n \sum_{k=0}^{n}\binom{n}{k}b^{k}c^{n-k}-b^{n} ∑k=0n(kn)bkcn−k−bn= ∑ k = 0 n − 1 ( n k ) b k c n − k \sum_{k=0}^{n-1}\binom{n}{k}b^{k}c^{n-k} ∑k=0n−1(kn)bkcn−k
- = c ∑ k = 0 n − 1 ( n k ) b k c n − k − 1 c\sum_{k=0}^{n-1}\binom{n}{k}b^{k}c^{n-k-1} c∑k=0n−1(kn)bkcn−k−1
- = ( a − b ) ∑ k = 0 n − 1 ( n k ) b k c n − k − 1 (a-b)\sum_{k=0}^{n-1}\binom{n}{k}b^{k}c^{n-k-1} (a−b)∑k=0n−1(kn)bkcn−k−1
- = ( a − b ) ∑ k = 1 n ( n k ) b k − 1 c n − k (a-b)\sum_{k=1}^{n}\binom{n}{k}b^{k-1}c^{n-k} (a−b)∑k=1n(kn)bk−1cn−k
- = ( a − b ) ∑ k = 1 n ( n k ) b k − 1 ( a − b ) n − k (a-b)\sum_{k=1}^{n}\binom{n}{k}b^{k-1}(a-b)^{n-k} (a−b)∑k=1n(kn)bk−1(a−b)n−k
- ⋯ \cdots ⋯(TODO)
- = ( a − b ) ∑ k = 1 n b n − k a k − 1 (a-b)\sum_{k=1}^{n}b^{n-k}a^{k-1} (a−b)∑k=1nbn−kak−1
等幂和
奇次方幂之和
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对于两个奇次方幂之和 a n + b n a^{n}+b^{n} an+bn, n = 1 , 3 , 5 , ⋯ , 2 k + 1 n={1,3,5,\cdots,2k+1} n=1,3,5,⋯,2k+1,可以按如下有
n次方数之差推导:-
n n n为奇数的情况下
- ( − b ) n = ( − 1 ) n b n = − b n (-b)^n=(-1)^nb^n=-b^n (−b)n=(−1)nbn=−bn
- a n − ( − b ) n = a n − ( − b n ) = a n + b n a^n-(-b)^n=a^n-(-b^n)=a^n+b^n an−(−b)n=an−(−bn)=an+bn
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a n + b n a^n+b^n an+bn= a n − ( − b ) n a^n-(-b)^n an−(−b)n,即可将 b 取 − b b取-b b取−b带入到等幂差公式中得到奇数等幂和公式
- 其中,a的指数与b的指数之和为
n-1
- 其中,a的指数与b的指数之和为
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例如, n = 3 n=3 n=3, a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3+b^3=(a+b)(a^2-ab+b^2) a3+b3=(a+b)(a2−ab+b2)
偶次方幂之和
- 偶次方幂之和的因式分解展开比较奇次方幂不容易展开
- 例如 x 2 + y 2 x^2+y^2 x2+y2在实数范围内无法因式分解,复数域内: x 2 + y 2 = ( x + y i ) ( x − y i ) x^2+y^2=(x+yi)(x-yi) x2+y2=(x+yi)(x−yi)
证明等幂差
- 不从推导的角度,从结论的角度证明等幂差展开公式的正确性
展开证明
- ( a − b ) ∑ i = 1 n a n − i b i − 1 (a-b) \sum_{i=1}^{n} a^{n-i}b^{i-1} (a−b)∑i=1nan−ibi−1, ( n ∈ N + ) {(n\in\mathbb{N^+})} (n∈N+)
- = a ∑ i = 1 n a n − i b i − 1 a\sum_{i=1}^{n} a^{n-i}b^{i-1} a∑i=1nan−ibi−1- b ∑ i = 1 n a n − i b i − 1 b\sum_{i=1}^{n} a^{n-i}b^{i-1} b∑i=1nan−ibi−1
- = ∑ i = 1 n a n − i + 1 b i − 1 \sum_{i=1}^{n} a^{n-i+1}b^{i-1} ∑i=1nan−i+1bi−1- ∑ i = 1 n a n − i b i \sum_{i=1}^{n} a^{n-i}b^{i} ∑i=1nan−ibi
- = ∑ i = 1 n a n − ( i − 1 ) b i − 1 \sum_{i=1}^{n} a^{n-(i-1)}b^{i-1} ∑i=1nan−(i−1)bi−1- ( ∑ i = 1 n − 1 a n − i b i + a 0 b n ) (\sum_{i=1}^{n-1} a^{n-i}b^{i}+a^{0}b^{n}) (∑i=1n−1an−ibi+a0bn)
- = ∑ i = 0 n a n − i b i \sum_{i=0}^{n} a^{n-i}b^{i} ∑i=0nan−ibi- ∑ i = 1 n − 1 a n − i b i \sum_{i=1}^{n-1} a^{n-i}b^{i} ∑i=1n−1an−ibi- a 0 b n a^{0}b^{n} a0bn
- = a n b 0 a^{n}b^{0} anb0+ ∑ i = 1 n a n − i b i \sum_{i=1}^{n} a^{n-i}b^{i} ∑i=1nan−ibi- ∑ i = 1 n − 1 a n − i b i \sum_{i=1}^{n-1} a^{n-i}b^{i} ∑i=1n−1an−ibi- a 0 b n a^{0}b^{n} a0bn
- = a n − b n a^{n}-b^{n} an−bn
ref
- 等幂和差 (wikipedia.org)