1,1,2,3,5,8,13,21,34,55.....的通项公式是什么?

Question

如题所述

Answer 1

。。裴波那契数列递推公式：F(n+2) = F(n+1) + F(n)F(1)=F(2)=1。它的通项求解如下：F(n+2) = F(n+1) + F(n) = F(n+2) - F(n+1) - F(n) = 0令 F(n+2) - aF(n+1) = b(F(n+1) - aF(n))展开 F(n+2) - (a+b)F(n+1) + abF(n) = 0显然 a+b=1 ab=-1由韦达定理知 a、b为二次方程 x^2 - x - 1 = 0 的两个根解得 a = (1 + √5)/2，b = (1 -√5)/2 或 a = (1 -√5)/2，b = (1 + √5)/2令G(n) = F(n+1) - aF(n)，则G(n+1) = bG(n)，且G(1) = F(2) - aF(1) = 1 - a = b，因此G(n)为等比数列，G(n) = b^n ，即F(n+1) - aF(n) = G(n) = b^n --------(1)在(1)式中分别将上述 a b的两组解代入，由于对称性不妨设x = (1 + √5)/2，y = (1 -√5)/2，得到：F(n+1) - xF(n) = y^nF(n+1) - yF(n) = x^n以上两式相减得：