n乗された2つの数の和を基本対称式で表す

n乗された2つの数の和を基本対称式で表す

$\alpha ,\beta$を解とする2次方程式を考える． $$\begin{array}{rcl}(x-\alpha )(x-\beta )& =& 0\\ x^2-(\alpha +\beta )x+\alpha \beta & =& 0\end{array}$$ $\alpha ,\beta$は式の解なので，$x$に代入しても式は成り立つ．代入した2つの式を得る． $$\begin{array}{r}\{\begin{array}{lll}{\alpha }^2-(\alpha +\beta )\alpha +\alpha \beta & =& 0\\ {\beta }^2-(\alpha +\beta )\beta +\alpha \beta & =& 0\end{array}\end{array}$$ 第一式には${\alpha }^{n-2}$，第二式には${\beta }^{n-2}$を両辺に掛ける． $$\begin{array}{r}\{\begin{array}{lll}{\alpha }^{n-2}\cdot \{{\alpha }^2-(\alpha +\beta )\alpha +\alpha \beta \}& =& {\alpha }^{n-2}\cdot 0\\ {\beta }^{n-2}\cdot \{{\beta }^2-(\alpha +\beta )\beta +\alpha \beta \}& =& {\beta }^{n-2}\cdot 0\end{array}\\ \\ \{\begin{array}{lll}{\alpha }^n-(\alpha +\beta ){\alpha }^{n-1}+\alpha \beta {\alpha }^{n-2}& =& 0\\ {\beta }^n-(\alpha +\beta ){\beta }^{n-1}+\alpha \beta {\beta }^{n-2}& =& 0\end{array}\end{array}$$ 両式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る． $$\begin{array}{rclrcl}& {\alpha }^n& -(\alpha +\beta ){\alpha }^{n-1}& +\alpha \beta {\alpha }^{n-2}& =& 0\\ +)& {\beta }^n& -(\alpha +\beta ){\beta }^{n-1}& +\alpha \beta {\beta }^{n-2}& =& 0\\ & {\alpha }^n+{\beta }^n& -(\alpha +\beta )({\alpha }^{n-1}+{\beta }^{n-1})& +\alpha \beta ({\alpha }^{n-2}+{\beta }^{n-2})& =& 0\end{array}$$ $$\begin{array}{rcl}{\alpha }^n+{\beta }^n& =& (\alpha +\beta )({\alpha }^{n-1}+{\beta }^{n-1})-\alpha \beta ({\alpha }^{n-2}+{\beta }^{n-2})\end{array}$$

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$\alpha ,\beta$は任意の数でよいのでそれを$x,y$とすれば以下の式を得る． $$\begin{array}{rcl}{x}^n+y^n& =& (x+y)(x^{n-1}+y^{n-1})-xy(x^{n-2}+y^{n-2})\end{array}$$

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問

original:https://www.youtube.com/watch?v=KPT862KhxRM $$\begin{array}{r}{\left(\frac{1+\sqrt{13}}{2}\right)}^7+{\left(\frac{1-\sqrt{13}}{2}\right)}^7\mathrm{の}\mathrm{値}\mathrm{を}\mathrm{求}\mathrm{め}\mathrm{よ}．\end{array}$$ 第一項,第二項の括弧内をそれぞれ$\alpha ,\beta$とする． $$\begin{array}{r}{\alpha }^7+{\beta }^7\cdots \alpha =\frac{1+\sqrt{13}}{2},\beta =\frac{1-\sqrt{13}}{2}\end{array}$$ n乗の和を基本対称式で求める上記式を用い，2乗から7乗まで順次求めていく． $$\begin{array}{rcl}{\alpha }^2+{\beta }^2& =& (\alpha +\beta )({\alpha }^{2-1}+{\beta }^{2-1})-\alpha \beta ({\alpha }^{2-2}+{\beta }^{2-2})\\ & =& (\alpha +\beta )(\alpha +\beta )-\alpha \beta ({\alpha }^0+{\beta }^0)\\ & =& 1\cdot 1-(-3)2=1+6=7\cdots \alpha +\beta =1,\alpha \beta =-3,{\alpha }^0+{\beta }^0=1+1=2\\ \\ {\alpha }^3+{\beta }^3& =& (\alpha +\beta )({\alpha }^{3-1}+{\beta }^{3-1})-\alpha \beta ({\alpha }^{3-2}+{\beta }^{3-2})\\ & =& (\alpha +\beta )({\alpha }^2+{\beta }^2)-\alpha \beta (\alpha +\beta )\\ & =& 1\cdot 7-(-3)1=7+3=10\\ \\ {\alpha }^4+{\beta }^4& =& (\alpha +\beta )({\alpha }^{4-1}+{\beta }^{4-1})-\alpha \beta ({\alpha }^{4-2}+{\beta }^{4-2})\\ & =& (\alpha +\beta )({\alpha }^3+{\beta }^3)-\alpha \beta ({\alpha }^2+{\beta }^2)\\ & =& 1\cdot 10-(-3)7=10+21=31\\ \\ {\alpha }^5+{\beta }^5& =& (\alpha +\beta )({\alpha }^{5-1}+{\beta }^{5-1})-\alpha \beta ({\alpha }^{5-2}+{\beta }^{5-2})\\ & =& (\alpha +\beta )({\alpha }^4+{\beta }^4)-\alpha \beta ({\alpha }^3+{\beta }^3)\\ & =& 1\cdot 31-(-3)10=31+30=61\\ \\ {\alpha }^6+{\beta }^6& =& (\alpha +\beta )({\alpha }^{6-1}+{\beta }^{6-1})-\alpha \beta ({\alpha }^{6-2}+{\beta }^{6-2})\\ & =& (\alpha +\beta )({\alpha }^5+{\beta }^5)-\alpha \beta ({\alpha }^4+{\beta }^4)\\ & =& 1\cdot 61-(-3)31=61+93=154\\ \\ {\alpha }^7+{\beta }^7& =& (\alpha +\beta )({\alpha }^{7-1}+{\beta }^{7-1})-\alpha \beta ({\alpha }^{7-2}+{\beta }^{7-2})\\ & =& (\alpha +\beta )({\alpha }^6+{\beta }^6)-\alpha \beta ({\alpha }^5+{\beta }^5)\\ & =& 1\cdot 154-(-3)61=154+183=337\end{array}$$