数列 $\{a_n\}$，$\{b_n\}$ 满足条件：$a_1=b_1=1$，$a_{n+1}=a_n+2b_n$，$b_{n+1}=a_n+ b_n$．证明：对每一个正整数 $n$，下式成立：
【难度】
【出处】
无
【标注】
-
$\dfrac {a_{2n-1}}{b_{2n-1}}<\sqrt 2$，$\dfrac {a_{2n }}{b_{2n }}>\sqrt 2$；标注答案略解析据条件$$a_{n+1}^2=2b_{b+1}^2=(a_n+2b_n)^2-2(a_n+b_n)^2=-(a_n^2-2b_n^2),\quad\cdots\cdots \text{ ① }$$得\[\begin{split} a_n^2-2b_n^2 &= -(a_{n-1}^2-2b_{n-1}^2)\\ &=(-1)^2(a_{n-2}^2-2b_{n-2}^2)\\&=\cdots \\&=(-1)^{n-1}(a_{1}^2-2b_{1}^2)=(-1)^n,\end{split}\]因此有$$a_{2n-1}^2-2b_{2n-1}^2<0 , a_{2n}^2-2b_{2n }^2>0,$$由条件知，$a_n$，$b_n$ 皆为正整数，因此得，$$\dfrac {a_{2n-1}}{b_{2n-1}}<\sqrt 2 , \dfrac {a_{2n}}{b_{2n}}>\sqrt 2.$$
-
$\left|\dfrac {a_{ n+1}}{b_{ n+1}}-\sqrt 2\right|<\left|\dfrac {a_{ n }}{b_{ n }}-\sqrt 2\right|$．标注答案略解析据 ① 得，$$\left|a_{n+1}^2-2b_{n+1}^2\right|=\left|a_{n }^2-2b_{n }^2\right|,\quad\cdots\cdots \text{ ② }$$又由$$a_{n+1}>a_n>0 , b_{n+1}>b_n>0,$$得到$$ \dfrac {1}{a_{n+1}+\sqrt 2b_{n+1}}<\dfrac {1}{a_{n }+\sqrt 2b_{n }},\quad\cdots\cdots \text{ ③ }$$$$\dfrac {1}{b_{n+1}}<\dfrac {1}{b_n}, \quad\cdots\cdots \text{ ④ }$$将 ②③④ 相乘得，$$\left|\dfrac {a_{n+1}}{b_{n+1}}-\sqrt 2\right|<\left|\dfrac {a_n}{b_n}-\sqrt 2\right|.$$
题目
问题1
答案1
解析1
备注1
问题2
答案2
解析2
备注2
