每日一题[704]数列“一锅鲜”

已知数列$\{a_n\}$是公差不为零的等差数列,$a_5=6$,数列$\{b_n\}$满足$b_1=3$,$b_{n+1}=b_1b_2\cdots b_n+1$.
(1) 当$n\geqslant 2$时,求证:$\dfrac{b_{n+1}-1}{b_n-1}=b_n$;
(2) 当$a_3>1$且$a_3\in\mathbb N^*$时,存在任意多项的等比数列$a_3,a_5,a_{k_1},a_{k_2},\cdots ,a_{k_n}$($n\in\mathbb N^*$),求$a_3$;
(3) 在(2)的条件下,当$a_3$取最小值时,求证:$$\dfrac{1}{b_1}+\dfrac{1}{b_2}+\cdots +\dfrac{1}{b_n}>4\left(\dfrac{1}{a_{k_1}-1}+\dfrac{1}{a_{k_2}-1}+\cdots +\dfrac{1}{a_{k_n}-1}\right).$$


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分析与解 (1) 根据题意,有$$\begin{cases} b_{n+1}-1=b_1b_2\cdots b_{n-1}b_n,\\b_{n}-1=b_1b_2\cdots b_{n-1},\end{cases} $$两式相除,于是原命题得证.

(2) 根据题意,有$$a_{k_n}=a_3\cdot\left(\dfrac{a_5}{a_3}\right)^{n+1}=a_3\cdot\left(\dfrac 6{a_3}\right)^{n+1},$$又$$a_{k_n}=a_3+\left(k_n-3\right)\cdot\dfrac{a_5-a_3}2=a_3+\left(k_n-3\right)\cdot \dfrac{6-a_3}2,$$因此可得$$k_n-3=2\cdot\dfrac{a_3\left[1-\left(\dfrac{6}{a_3}\right)^{n+1}\right]}{a_3-6}=2\sum_{i=0}^n\left(\dfrac{6}{a_3}\right)^i.$$当$n$分别取$1$和$2$时,有$$\begin{cases} k_1-3=2+\dfrac{12}{a_3},\\ k_2-3=2+\dfrac{12}{a_3}+\dfrac{72}{a_3^2},\end{cases} $$于是$a_3=2$或$a_3=3$.经验证,$a_3=2$和$a_3=3$都符合题意.因此$a_3$的所有可能取值为$2,3$.

(3) 根据题意,当$n\geqslant 2$时有\[\begin{split} LHS&=\sum_{i=1}^n\dfrac{1}{b_i}=\sum_{i=1}^n\left(\dfrac{1}{b_i-1}-\dfrac{1}{b_{i+1}-1}\right)\\&=\dfrac{1}{b_1}+\dfrac{1}{b_2}+\dfrac 1{b_3-1}-\dfrac{1}{b_{n+1}-1}\\&=\dfrac 23-\dfrac{1}{b_1b_2\cdots b_n}\geqslant \dfrac 23-\dfrac{1}{3^n}\geqslant \dfrac 59.\end{split} \]另一方面,由于$a_3=2$,于是$$RHS=\sum_{i=1}^n\dfrac{4}{2\cdot 3^{i+1}-1}<\dfrac {\frac 4{17}}{1-\frac 13}=\dfrac 6{17}<\dfrac 59.$$因此原命题得证.

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