每日一题[506]递推中的不变量

已知数列$\{a_n\}$和$\{b_n\}$满足$a_1=1$,$b_1=2$,$a_{n+1}b_n=a_nb_n+2a_n+4$.

(1)若$b_n=2a_n$,求证:当$n\geqslant 2$时,$n+2\leqslant a_n\leqslant \dfrac 32n+1$;

(2)若$a_nb_{n+1}=a_nb_n+2b_n+4$,求证:$a_n<10$.


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分析与解    (1)根据题意有$$a_{n+1}=a_n+\dfrac{2}{a_n}+1,$$显然有$\{a_n\}$单调递增.又$a_2=4$,因此当$n\geqslant 2$时,有$$1\leqslant a_{n+1}-a_n\leqslant \dfrac 2{a_2}+1=\dfrac 32,$$累加即得.

(2)由$a_{n+1}b_n=a_nb_n+2a_n+4$得$$b_n=\dfrac{2a_n+4}{a_{n+1}-a_n},$$于是$$\dfrac{a_n(2a_{n+1}+4)}{a_{n+2}-a_{n+1}}=\dfrac{(a_n+2)(2a_n+4)}{a_{n+1}-a_n}+4,$$即$$\dfrac{a_{n+1}+2}{a_{n+2}-a_{n+1}}=\dfrac{a_n^2+2a_n+2a_{n+1}+4}{a_n(a_{n+1}-a_n)},$$于是\[\begin{split} \dfrac{a_{n+1}+2}{(a_{n+2}-a_{n+1})+(a_{n+1}+2)}&=\dfrac{a_n^2+2a_n+2a_{n+1}+4}{(a_na_{n+1}-a_n^2)+(a_n^2+2a_n+2a_{n+1}+4)}\\ &=\dfrac{(a_n+2)^2+2(a_{n+1}+2)-2(a_n+2)}{(a_n+2)(a_{n+1}+2)}\\ &=\dfrac{a_n+2}{a_{n+1}+2}+\dfrac 2{a_n+2}-\dfrac{2}{a_{n+1}+2},\end{split} \]因此$$\dfrac{a_{n+1}+2}{a_{n+2}+2}+\dfrac{2}{a_{n+1}+2}=\dfrac{a_n+2}{a_{n+1}+2}+\dfrac{2}{a_n+2},$$因此数列$\left\{\dfrac{a_n+2}{a_{n+1}+2}+\dfrac{2}{a_n+2}\right\}$为常数列.

事实上,$a_1=1$,$a_2=4$,因此$$\dfrac{a_n+2}{a_{n+1}+2}+\dfrac{2}{a_n+2}=\dfrac 76,$$设$c_n=\dfrac{1}{a_n+2}$,则$$\dfrac{c_{n+1}}{c_n}+2c_n=\dfrac 76,$$即$$c_{n+1}=-2c_n^2+\dfrac 76c_n,$$利用不动点改造递推公式,得$$c_{n+1}-\dfrac 1{12}=\left(c_n-\dfrac 1{12}\right)\cdot (1-2c_n).$$显然$\{a_n\}$单调递增,于是$$c_n=\dfrac{1}{a_n+2}\leqslant \dfrac {1}{1+2}=\dfrac 13,$$因此$$\forall n\in\mathcal N^*,c_n-\dfrac 1{12}>0,$$原命题得证.

   用合分比定理的那一步取倒数亦可.

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