每日一题[1000]数列与不等式

设实数$a\geqslant 2$,方程$x^2-ax+1=0$的两根分别为$x_1,x_2$,$a_n=x_1^n+x_2^n$($n=1,2,\cdots$),$b_n=\dfrac{a_n}{a_{n+1}}$.
(1)用$b_n$表示$b_{n+1}$,并判断数列$\{b_n\}$的单调性;
(2)求所有实数$a$的值,使得对任意正整数$n$都有$b_1+b_2+\cdots+b_n>n-1$.


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分析与解        (1) 根据题意,不妨设$x_1=\alpha$,$x_2=\dfrac{1}{\alpha}$,$a=\alpha+\dfrac{1}{\alpha}$,且$\alpha\geqslant 1$,则\[\begin{split}a_{n+2}=&\alpha^{n+2}+\dfrac{1}{\alpha^{n+2}}\\=&\left(\alpha+\dfrac{1}{\alpha}\right)\cdot \left(\alpha^{n+1}+\dfrac{1}{\alpha^{n+1}}\right)-\left(\alpha^{n}+\dfrac{1}{\alpha^{n}}\right)\\=&a\cdot a_{n+1}-a_{n},\end{split}\]即\[\dfrac{a_{n+2}}{a_{n+1}}=a-\dfrac{a_{n}}{a_{n+1}},\]整理得\[b_{n+1}=\dfrac{1}{a-b_n}.\]因此\[\dfrac{b_{n+1}}{b_n}=\dfrac{1}{b_n(a-b_n)}.\]因为$\alpha\geqslant 1$,所以$$\dfrac 1{\alpha}\leqslant b_n=\dfrac {a_n}{a_{n+1}}=\dfrac {\alpha^n+\dfrac 1{\alpha^n}}{\alpha^{n+1}+\dfrac 1{\alpha^{n+1}}}\leqslant \alpha,$$于是\[b_n(a-b_n)\geqslant 1,\]等号当且仅当$a=2$(此时$\alpha=1$)时取得,因此数列$\{b_n\}$当$a=2$时为常数列,当$a>2$时单调递减.

(2) 当$a>2$时,数列$\{b_n\}$单调递减趋于$\dfrac{1}{\alpha}$,于是必然存在$\varepsilon>0$,使得当$n>N$时,均有\[b_n-\dfrac 1{\alpha}<\varepsilon,\]于是\[\begin{split}LHS&=\left(b_1+b_2+\cdots+b_N\right)+\left(b_{N+1}+\cdots+b_n\right)\\&<N\alpha+(n-N)\left(\dfrac{1}{\alpha}+\varepsilon\right)\\&=\dfrac{1}{\alpha}\cdot n+N\left(\alpha-\dfrac{1}{\alpha}\right)-N\varepsilon.\end{split}\]考虑到$\dfrac{1}{\alpha}<1$,因此必然存在$M$,使得当$n>M$时,有\[\dfrac{1}{\alpha}\cdot n+N\left(\alpha-\dfrac{1}{\alpha}\right)-N\varepsilon<n-1,\]不符合题意.

当$a=2$时,数列$\{b_n\}$为常数列$b_n=1$($n\in\mathbb N^*$),此时显然符合题意.

综上所述,所有满足要求的实数$a$的值为$2$.

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每日一题[1000]数列与不等式》有 2 条评论

  1. cbc123e说:

    为什么
    1/alpha <= bn <= alpha?

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