Question

斐波那契数列通向公式的问题设常数r,s.使得F(n)-r*F(n-1)=s*[F(n-1)-r*F(n-2)].则r+s

Answer 1

n>=3时,

f(n) - rf(n-1) = s[f(n-1) - rf(n-2)]

n>=1时,

f(n+2) - rf(n+1) = s[f(n+1) -rf(n)],

{f(n+1)-rf(n)}是首项为f(2)-rf(1)=1-r=s, 公比为s的等比数列.

f(n+1)-rf(n) = s*s^(n-1) = s^n,

n>=2时,

f(n) - rf(n-1) = s^(n-1),

f(n) = s^(n-1) + rf(n-1).

这样更好理解吧~~