每日一题[702]积极改造

（2010年安徽卷）设$a_1,a_2,\cdots ,a_n,\cdots$中的每一项均不为$0$，求证：$\{a_n\}$是等差数列的充分必要条件是对任意自然数$n$，均有

$$\dfrac{1}{a_1a_2}+\dfrac 1{a_2a_3}+\cdots +\dfrac{1}{a_na_{n+1}}=\dfrac{n}{a_1a_{n+1}}.$$

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分析与证明　先证必要性
若$\{a_n\}$是公差为$d$的等差数列，则

$$\dfrac{1}{a_ka_{k+1}}=\dfrac 1d\cdot \dfrac{a_{k+1}-a_k}{a_ka_{k+1}}=-\dfrac 1d\left(\dfrac{1}{a_{k+1}}-\dfrac{1}{a_k}\right),$$ 所以 $$\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots +\dfrac{1}{a_na_{n+1}}=-\dfrac 1d\left(\dfrac 1{a_{n+1}}-\dfrac 1{a_1}\right)=\dfrac{n}{a_1a_{n+1}}.$$

再证充分性
根据题意，有

$$\begin{cases} \dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots +\dfrac{1}{a_na_{n+1}}=\dfrac{n}{a_1a_{n+1}},\\ \dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots +\dfrac{1}{a_na_{n+1}}+\dfrac{1}{a_{n+1}a_{n+2}}=\dfrac{n+1}{a_1a_{n+2}},\end{cases}$$ 两式相减，得 $$\dfrac{1}{a_{n+1}a_{n+2}}=\dfrac{n+1}{a_1a_{n+2}}-\dfrac{n}{a_1a_{n+1}},$$ 即 $$a_1=(n+1)a_{n+1}-na_{n+2},$$ 也即 $$\dfrac{a_{n+1}-a_1}{n}=\dfrac{a_{n+2}-a_1}{n+1},$$ （注　也可以再写一项 $$a_1=na_n-(n-1)a_{n+1},$$ 两式相减得到 $$2a_{n+1}-a_{n+2}-a_n=0.$$ 从而证明是等差数列）

依次类推，有

$$\dfrac{a_{n+2}-a_1}{n+1}=\dfrac{a_{n+1}-a_1}{n}=\dfrac{a_n-a_1}{n-1}=\cdots =\dfrac{a_2-a_1}{1},$$ 于是 $$a_{n+1}=n(a_2-a_1)+a_1,$$ 因此数列$\{a_n\}$是等差数列．

综上所述，原命题得证．

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思考与总结　根据要证明的结论，对递推式进行恰当的改造．