Question

2,5,9,13,19 以此类推 这组数列的通项公式是什么

Answer 1

斐波那契数列,这么著名的数列都不知道!用特征方程法加待定系数法!

解 :设an-αa(n-1)=β(a(n-1)-αa(n-2))

得α+β=1

αβ=-1

构造方程x²-x-1=0,解得α=(1-√5)/2,β=(1+√5)/2或α=(1+√5)/2,β=(1-√5)/2

所以

an-(1-√5)/2*a(n-1)=(1+√5)/2*(a(n-1)-(1-√5)/2*a(n-2))=[(1+√5)/2]^(n-2)*(a2-(1-√5)/2*a1)`````````1

an-(1+√5)/2*a(n-1)=(1-√5)/2*(a(n-1)-(1+√5)/2*a(n-2))=[(1-√5)/2]^(n-2)*(a2-(1+√5)/2*a1)`````````2

由式1,式2,可得

an=[(1+√5)/2]^(n-2)*(a2-(1-√5)/2*a1)``````````````3

an=[(1-√5)/2]^(n-2)*(a2-(1+√5)/2*a1)``````````````4

将式3*(1+√5)/2-式4*(1-√5)/2,化简得an=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}