斐波那契数列通项公式的推导

(F_n=F_{n-1}+F_{n-2})

(frac{1}{-k}=frac{1-k}{1})

(k=frac{1+sqrt{5}}{2})

(F_n-frac{1+sqrt{5}}{2}F_{n-1}=frac{1-sqrt{5}}{2}(F_{n-1}-frac{1+sqrt{5}}{2}F_{n-2}))

(T_n=F_n-frac{1+sqrt{5}}{2}F_{n-1}=frac{1-sqrt{5}}{2}T_{n-1})

(T_1=1,T_n=(frac{1-sqrt{5}}{2})^{n-1})

(F_{n+1}=frac{1+sqrt{5}}{2}F_n+(frac{1-sqrt{5}}{2})^n)

(F_{n+1}=sum_{i=0}^n(frac{1-sqrt{5}}{2})^i(frac{1+sqrt{5}}{2})^{n-i})

(F_{n+1}=sum_{i=0}^n (-1)^i(frac{1+sqrt{5}}{2})^{n-2i})

(q=-(frac{1-sqrt{5}}{2})^2=frac{sqrt{5}-3}{2})

(F_{n+1}=frac{2}{5-sqrt{5}}[(frac{1-sqrt{5}}{2})^{n+2}+(frac{1+sqrt{5}}{2})^n])

(F_{n+1}=frac{sqrt{5}}{5}[(frac{1+sqrt{5}}{2})^{n+1}-(frac{1-sqrt{5}}{2})^{n+1}])

(F_n=frac{sqrt{5}}{5}[(frac{1+sqrt{5}}{2})^n-(frac{1-sqrt{5}}{2})^n])