问题描述：
斐波那契数列通项公式的几种求法
问题解答：
我来补答
1.x(1) = 1, x(2) = 1, x(3) = x(1) + x(2) = 2, ..., x(n) = x(n-1) + x(n-2), ...这就是斐波那契数列
设x(n) + a1 x(n-1) = a2(x(n-1) + a1 x(n-2))a2 - a1 = 1a2 X a1 = 1
{x(n) + a1 x(n-1)} 就是等比数列结果为 x(n) + a1 x(n-1) = b1 X a2^n设x(n) + c1 X a2^n = c2 (x(n-1) + c1 X a2^(n-1))c2 = -a1c1 X c2 / a2 - c1 = b1
{x(n) + c1 X a2^n}为等比数列计算出上面的所有待定的参数, 就容易得到:
设x(n) + a1 x(n-1) = a2(x(n-1) + a1 x(n-2))a2 - a1 = 1a2 X a1 = 1
{x(n) + a1 x(n-1)} 就是等比数列结果为 x(n) + a1 x(n-1) = b1 X a2^n设x(n) + c1 X a2^n = c2 (x(n-1) + c1 X a2^(n-1))c2 = -a1c1 X c2 / a2 - c1 = b1
{x(n) + c1 X a2^n}为等比数列计算出上面的所有待定的参数, 就容易得到:
展开全文阅读