(清华大学THUSSAT)
已知 (a=left( dfrac{-1+sqrt{5}}{2}
ight)^{-10}+left( dfrac{-1-sqrt{5}}{2}
ight)^{-10}, b=left( dfrac{-1+sqrt{5}}{2}
ight)^{10}+left( dfrac{-1-sqrt{5}}{2}
ight)^{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)