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Giả sử: gcd$(kF_{n+2}+F_n;kF_{n+3}+F_{n+1})=d$
$\Rightarrow \left\{ \begin{array}{l} d\,|\,kF_{n+3}+F_{n+1}-kF_{n+2}-F_n=kF_{n+1}+F_{n-1}\\ d\,|\,kF_{n+3}+F_{n+1}+kF_{n+2}+F_n=kF_{n+4}+F_{n+2} \end{array} \right.$
Bằng quy nạp suy ra: $d\,|\,kF_{n+2}+F_n,\forall n\in\mathbb{N}$
Suy ra:
$\left\{ \begin{array}{l} d\,|\,kF_2+F_0=2k+1\\d\,|\,kF_3+F_1=3k+1 \end{array} \right.\Rightarrow \left\{ \begin{array}{l} d\,|\,6k+3\\d\,|\,6k+2 \end{array} \right.$
$\Rightarrow d\,|\,(6k+3)-(6k+2)=1\Rightarrow d=1$.
Suy ra: $\frac{kF_{n+2}+F_n}{kF_{n+3}+F_{n+1}}$ tối giản.
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Trả lời 12-11-12 09:34 PM
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