MT【141】逆用特征根法

(清华大学THUSSAT)
已知 \(a=\left( \dfrac{-1+\sqrt{5}}{2} \right)^{-10}+\left( \dfrac{-1-\sqrt{5}}{2} \right)^{-10},\ b=\left( \dfrac{-1+\sqrt{5}}{2} \right)^{10}+\left( \dfrac{-1-\sqrt{5}}{2} \right)^{10}\),则点 \(P(a,b)\) 的坐标为_____

解答:
显然\(a=b\),设$ x_1=\dfrac{-1+\sqrt{5}}{2}, x_2= \dfrac{-1-\sqrt{5}}{2} $ ,则 \(x_1+x_2=-1,x_1x_2=-1\)
\(x_1,x_2\)为方程\(x^2+x-1=0\)的两根.此特征方程对应的递推数列为\(a_{n+1}=a_{n-1}-a_n\)
易知\(a_1=-1,a_2=3\)故有递推式知:\(a_3=-4,a_4=7,a_5=-11,a_6=18,a_7=-29,a_8=47,a_9=-76,a_{10}=123\),显然\(a=b=a_{10}=123\)