（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$$\dfrac{1}{a_1a_2}+\dfrac 1{a_2a_3}+\cdots +\dfrac{1}{a_na_{n+1}}=\dfrac{n}{a_1a_{n+1}}.$$
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．
思考与总结 根据要证明的结论，对递推式进行恰当的改造．