n乗同士の和と差の因数分解
n乗同士の和と差の因数分解
\(a,b\in\mathbb{R}\;,\;n\in\mathbb{N}_{0}\)とする
\(a,b\in\mathbb{R}\;,\;n\in\mathbb{N}_{0}\)とする
(1)
\[ a^{n}-b^{n}=\left(a-b\right)\left(\sum_{k=0}^{n-1}a^{n-1-k}b^{k}\right) \](2)
\[ a^{2n+1}\pm b^{2n+1}=\left(a\pm b\right)\left(\sum_{k=0}^{2n}\left(\mp1\right)^{k}a^{2n-k}b^{k}\right) \](1)
\begin{align*} a^{n}-b^{n} & =a^{n}+\left(\sum_{k=1}^{n-1}a^{n-k}b^{k}\right)-\left(\sum_{k=1}^{n-1}a^{n-k}b^{k}\right)-b^{n}\\ & =\left(\sum_{k=0}^{n-1}a^{n-k}b^{k}\right)-\left(\sum_{k=1}^{n}a^{n-k}b^{k}\right)\\ & =a\left(\sum_{k=0}^{n-1}a^{n-1-k}b^{k}\right)-b\left(\sum_{k=1}^{n}a^{n-k}b^{k-1}\right)\\ & =a\left(\sum_{k=0}^{n-1}a^{n-1-k}b^{k}\right)-b\left(\sum_{k=0}^{n-1}a^{n-1-k}b^{k}\right)\\ & =\left(a-b\right)\left(\sum_{k=0}^{n-1}a^{n-1-k}b^{k}\right) \end{align*}(2)
\begin{align*} a^{2n+1}\pm b^{2n+1} & =a^{2n+1}+\left(\sum_{k=1}^{2n}\left(\mp1\right)^{k}a^{2n+1-k}b^{k}\right)-\left(\sum_{k=1}^{2n}\left(\mp1\right)^{k}a^{2n+1-k}b^{k}\right)\pm b^{2n+1}\\ & =\left(\sum_{k=0}^{2n}\left(\mp1\right)^{k}a^{2n+1-k}b^{k}\right)-\left(\sum_{k=1}^{2n+1}\left(\mp1\right)^{k}a^{2n+1-k}b^{k}\right)\\ & =a\left(\sum_{k=0}^{2n}\left(\mp1\right)^{k}a^{2n-k}b^{k}\right)-b\left(\sum_{k=1}^{2n+1}\left(\mp1\right)^{k}a^{2n+1-k}b^{k-1}\right)\\ & =a\left(\sum_{k=0}^{2n}\left(\mp1\right)^{k}a^{2n-k}b^{k}\right)-b\left(\mp\sum_{k=0}^{2n+1}\left(\mp1\right)^{k}a^{2n-k}b^{k}\right)\\ & =\left(a\pm b\right)\left(\sum_{k=0}^{2n}\left(\mp1\right)^{k}a^{2n-k}b^{k}\right) \end{align*}ページ情報
| タイトル | n乗同士の和と差の因数分解 |
| URL | https://www.nomuramath.com/ddv6jkbk/ |
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3次方程式の標準形
\[
X^{3}+pX+q=0
\]
ビネ・コーシーとラグランジュの恒等式
\[
\left(\sum_{i=1}^{n}a_{i}c_{i}\right)\left(\sum_{j=1}^{n}b_{j}d_{j}\right)-\left(\sum_{i=1}^{n}a_{i}d_{i}\right)\left(\sum_{j=1}^{n}b_{j}c_{j}\right)=\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(c_{i}d_{j}-c_{j}d_{i}\right)
\]
3次式の実数の範囲で因数分解
\[
a^{3}\pm b^{3}=\left(a\pm b\right)\left(a^{2}\mp ab+b^{2}\right)
\]
差積の定義
\[
\Delta\left(x_{1},\cdots,x_{n}\right):=\prod_{1\leq i<j\leq n}\left(x_{i}-x_{j}\right)
\]