题目（高二 · 代数）：20. 已知数列 $\left\{a_{n}\right\},\left\{b_{n}\right\}$ 均为各项都不相等的数列, $S_{n}$ 为 $\left\{a_{n}\right\}$ 的前 $n$ 项和, $a_{n+1} b_{n}=S_{n}+1\left(n \in N^{*}\right)$.

(1) 若 $a_{1}=1, b_{n}=\frac{n}{2}$, 求 $a_{4}$ 的值;

(2) 若 $\left\{a_{n}\right\}$ 是公比为 $q(q \neq 1)$ 的等比数列, 求证: 数列 $\left\{b_{n}+\frac{1}{1-q}\right\}$ 为等比数列;

(3) 若 $\left\{a_{n}\right\}$ 的各项都不为零, $\left\{b_{n}\right\}$ 是公差为 $d$ 的等差数列, 求证: $a_{2}, a_{3}, \cdots, a_{n}, \cdots$ 成等差数列的充要条件是 $d=\frac{1}{2}$.
选项：
答案：(1) $\because a_{n+1} b_{n}=S_{n}+1, \quad a_{1}=1, \quad b_{n}=\frac{n}{2}$,

$\therefore a_{2}=\frac{a_{1}+1}{b_{1}}=\frac{1+1}{\frac{1}{2}}=4, \frac{1+4+6+1}{\frac{3}{2}}$
$a_{3}=\frac{S_{2}+1}{b_{2}}=\frac{1+4+1}{1}=6$,

$a_{4}=\frac{S_{3}+1}{b_{3}}=\frac{1+4+6+1}{\frac{3}{2}}=8$,

证明: (2) 设 $a_{n}=a_{1} q^{n-1}(q \neq 1)$, 则 $S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}$,

$\because a_{n+1} b_{n}=S_{n}+1$,

$\therefore b_{n}=\frac{S_{n}+1}{a_{n+1}}=\frac{\frac{a_{1}\left(1-q^{n}\right)}{1-q}+1}{a_{1} q^{n}}=\frac{a_{1}-a_{1} q^{n}+1-q}{(1-q) a_{1} q^{n}}$,

$\therefore b_{n}+\frac{1}{1-q}=\frac{a_{1}-a_{1} q^{n}+1-q}{(1-q) a_{1} q^{n}}+\frac{1}{1-q}=\frac{a_{1}+1-q}{(1-q) a_{1} q^{n}}$

$\therefore b_{n+1}+\frac{1}{1-q}=\frac{a_{1}+1-q}{(1-q) a_{1} q^{n+1}}$

$\therefore \frac{b_{n+1}}{b_{n}}=q,(q$ 为常数)

$\therefore$ 数列 $\left\{b_{n}+\frac{1}{1-q}\right\}$ 为等比数列,

(3) $\because$ 数列 $\left\{b_{n}\right\}$ 是公差为 $\mathrm{d}$ 的等差数列,

$\therefore$ 当 $n \geq 2$ 时, $a_{n+1} b_{n}-a_{n}\left(b_{n}-d\right)=a_{n}$,

即 $\left(a_{n+1}-a_{n}\right) b_{n}=(1-d) a_{n}$,

$\because$ 数列 $\left\{a_{n}\right\}$ 的各项都不为零,

$\therefore a_{n+1}-a_{n} \neq 0, \quad 1-d \neq 0$,

$\therefore$ 当 $n \geq 2$ 时, $\frac{b_{n}}{1-d}=\frac{a_{n}}{a_{n+1}-a_{n}}$,

当 $n \geq 3$ 时, $\frac{b_{n-1}}{1-d}=\frac{a_{n-1}}{a_{n}-a_{n-1}}$,

两式相减得: 当 $n \geq 3$ 时, $\frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=\frac{b_{n}-b_{n-1}}{1-d}=\frac{d}{1-d}$.
先证充分性:

由 $d=\frac{1}{2}$ 可知 $\frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=1$,

$\therefore$ 当 $n \geq 3$ 时, $\frac{a_{n-1}}{a_{n}-a_{n-1}}+1=\frac{a_{n}}{a_{n+1}-a_{n}}$,

又 $\because a_{n} \neq 0$,

$\therefore a_{n+1}-a_{n}=a_{n}-a_{n-1}$,

即 $a_{2}, a_{3}, \cdots, a_{n} \cdots$ 成等差数列;

再证必要性:

$\because a_{2}, a_{3}, \cdots, a_{n} \cdots$ 成等差数列,

$\therefore$ 当 $n \geq 3$ 时, $a_{n+1}-a_{n}=a_{n}-a_{n-1}$,

$\therefore \frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=\frac{a_{n-1}}{a_{n}-a_{n-1}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=1=\frac{d}{1-d}$,

$\therefore d=\frac{1}{2}$.

综上所述, $a_{2}, a_{3}, \cdots, a_{n} \ldots$ 成等差数列的充要条件是 $d=\frac{1}{2}$ 。
解析：(1) $\because a_{n+1} b_{n}=S_{n}+1, \quad a_{1}=1, \quad b_{n}=\frac{n}{2}$,

$\therefore a_{2}=\frac{a_{1}+1}{b_{1}}=\frac{1+1}{\frac{1}{2}}=4, \frac{1+4+6+1}{\frac{3}{2}}$
$a_{3}=\frac{S_{2}+1}{b_{2}}=\frac{1+4+1}{1}=6$,

$a_{4}=\frac{S_{3}+1}{b_{3}}=\frac{1+4+6+1}{\frac{3}{2}}=8$,

证明: (2) 设 $a_{n}=a_{1} q^{n-1}(q \neq 1)$, 则 $S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}$,

$\because a_{n+1} b_{n}=S_{n}+1$,

$\therefore b_{n}=\frac{S_{n}+1}{a_{n+1}}=\frac{\frac{a_{1}\left(1-q^{n}\right)}{1-q}+1}{a_{1} q^{n}}=\frac{a_{1}-a_{1} q^{n}+1-q}{(1-q) a_{1} q^{n}}$,

$\therefore b_{n}+\frac{1}{1-q}=\frac{a_{1}-a_{1} q^{n}+1-q}{(1-q) a_{1} q^{n}}+\frac{1}{1-q}=\frac{a_{1}+1-q}{(1-q) a_{1} q^{n}}$

$\therefore b_{n+1}+\frac{1}{1-q}=\frac{a_{1}+1-q}{(1-q) a_{1} q^{n+1}}$

$\therefore \frac{b_{n+1}}{b_{n}}=q,(q$ 为常数)

$\therefore$ 数列 $\left\{b_{n}+\frac{1}{1-q}\right\}$ 为等比数列,

(3) $\because$ 数列 $\left\{b_{n}\right\}$ 是公差为 $\mathrm{d}$ 的等差数列,

$\therefore$ 当 $n \geq 2$ 时, $a_{n+1} b_{n}-a_{n}\left(b_{n}-d\right)=a_{n}$,

即 $\left(a_{n+1}-a_{n}\right) b_{n}=(1-d) a_{n}$,

$\because$ 数列 $\left\{a_{n}\right\}$ 的各项都不为零,

$\therefore a_{n+1}-a_{n} \neq 0, \quad 1-d \neq 0$,

$\therefore$ 当 $n \geq 2$ 时, $\frac{b_{n}}{1-d}=\frac{a_{n}}{a_{n+1}-a_{n}}$,

当 $n \geq 3$ 时, $\frac{b_{n-1}}{1-d}=\frac{a_{n-1}}{a_{n}-a_{n-1}}$,

两式相减得: 当 $n \geq 3$ 时, $\frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=\frac{b_{n}-b_{n-1}}{1-d}=\frac{d}{1-d}$.
先证充分性:

由 $d=\frac{1}{2}$ 可知 $\frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=1$,

$\therefore$ 当 $n \geq 3$ 时, $\frac{a_{n-1}}{a_{n}-a_{n-1}}+1=\frac{a_{n}}{a_{n+1}-a_{n}}$,

又 $\because a_{n} \neq 0$,

$\therefore a_{n+1}-a_{n}=a_{n}-a_{n-1}$,

即 $a_{2}, a_{3}, \cdots, a_{n} \cdots$ 成等差数列;

再证必要性:

$\because a_{2}, a_{3}, \cdots, a_{n} \cdots$ 成等差数列,

$\therefore$ 当 $n \geq 3$ 时, $a_{n+1}-a_{n}=a_{n}-a_{n-1}$,

$\therefore \frac{a_{n}}{a_{n+1}-a_{n}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=\frac{a_{n-1}}{a_{n}-a_{n-1}}-\frac{a_{n-1}}{a_{n}-a_{n-1}}=1=\frac{d}{1-d}$,

$\therefore d=\frac{1}{2}$.

综上所述, $a_{2}, a_{3}, \cdots, a_{n} \ldots$ 成等差数列的充要条件是 $d=\frac{1}{2}$ 。