斐波拉契数列问题有一组斐波拿契数列1,1,2,3,5,8,13,21,34,55,89……问第N个数是多少用带N的式子列

F(n)= (1/√5){[(1+√5)/2]^(n+2)-[(1-√5)/2]^(n+2)}
下面用特征值法求F(n)——裴波那契数列 1 1 2 3 5 ...的通项
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 x2 - 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) = (1-a)bn-1 = bn ,即
F(n+1) - aF(n) = G(n) = bn -------------------------------------- (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) = yn
F(n+1) - yF(n) = xn
以上两式相减得：
(x-y)F(n) = xn - yn
F(n) = (xn - yn)/(x-y) = {[(1+√5)/2]n-[(1-√5)/2] n}/√5

结果懒得写了，告诉你过程
f(n) = f(n-1) + f(n)
设f(n) - af(n-1) = b(f(n-1) - af(n-2))
则a+b=1 a*b=-1 可以求出a,b
又设g(n) = f(n + 1) - af(n)为等比数列g(n) = bg(n-1)且g(1) = 1-a = b
所以g(n) = b^n
即f(n + 1)...

A1=1,A2=1,A3=2所以A(n+2)=A(n+1)+A(n)
A1=1
A2=1
A3=A2+A1
A4=A3+A2
…………
A(n-1)=A(n-2)+A(n-3)
...