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
\[
X^{3}+pX+q=0
\]
オイラーの4平方恒等式
\[
\left(a_{0}^{\;2}+a_{1}^{\;2}+a_{2}^{\;2}+a_{3}^{\;2}\right)\left(b_{0}^{\;2}+b_{1}^{\;2}+b_{2}^{\;2}+b_{3}^{\;2}\right)=\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)^{2}+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)^{2}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)^{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)^{2}
\]
4次方程式標準形の解き方
\[
y=\frac{\mp_{1}\sqrt{2u-p}\pm_{2}\sqrt{-p-2u-\frac{4q}{2\sqrt{2u-p}}}}{2}
\]
2次式の実数の範囲で因数分解
\[
a^{2}\pm2ab+b^{2}=\left(a\pm b\right)^{2}
\]