回復 5# satsuki931000 的帖子
第 7 題
\begin{align}
& \alpha +\beta =1 \\
& \alpha \beta =-1 \\
& \\
& {{\alpha }^{n+1}}-{{\beta }^{n+1}}=\left( {{\alpha }^{n}}-{{\beta }^{n}} \right)\left( \alpha +\beta \right)-\alpha \beta \left( {{\alpha }^{n-1}}-{{\beta }^{n-1}} \right)=\left( {{\alpha }^{n}}-{{\beta }^{n}} \right)+\left( {{\alpha }^{n-1}}-{{\beta }^{n-1}} \right) \\
& {{\alpha }^{2019}}-{{\beta }^{2019}}=\frac{3m+n}{2} \\
\end{align}
