设数列 ${a_1}$，${a_2}$，$\cdots$，${a_n}$，$\cdots $ 中的每一项都不为 $ 0 $．证明：$\left\{ {a_n}\right\} $ 为等差数列的充分必要条件是：对任何 $n \in{\mathbb{ N}}^*$，都有 $\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}} = \dfrac{n}{{{a_1}{a_{n + 1}}}}$．
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【解析】
（i）必要性
设数列 $\left\{ {a_n}\right\}$ 的公差为 $d$，若 $d = 0$，则所述等式显然成立；
若 $d \ne 0$，则\[\dfrac{1}{a_{k}a_{k+1}}=\dfrac{1}{d}\cdot \dfrac{a_{k+1}-a_{k}}{a_{k}a_{k+1}}=-\dfrac{1}{d}\left(\dfrac{1}{a_{k+1}}-\dfrac{1}{a_{k}}\right),\]所以\[\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}}= -\dfrac{1}{d}\left( \dfrac{1}{{{a_{n + 1}}}}-\dfrac{1}{a_1}\right) = \dfrac{1}{d} \cdot\dfrac{{{a_{n + 1}} - {a_1}}}{{{a_1}{a_{n + 1}}}}= \dfrac{n}{{{a_1}{a_{n + 1}}}}.\]（ii）充分性
依题意有\[\begin{split}&\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}} = \dfrac{n}{{{a_1}{a_{n + 1}}}}, \\& \dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}} + \dfrac{1}{{{a_{n + 1}}{a_{n + 2}}}} = \dfrac{n + 1}{{{a_1}{a_{n + 2}}}}. \end{split}\]两式相减得\[\dfrac{1}{{{a_{n + 1}}{a_{n + 2}}}} = \dfrac{n + 1}{{{a_1}{a_{n + 2}}}} - \dfrac{n}{{{a_1}{a_{n + 1}}}},\]即\[{a_1} = \left(n + 1\right){a_{n + 1}} - n{a_{n + 2}}\]从而可得\[{a_1} = n{a_n} - \left(n - 1\right){a_{n + 1}}, \]因此对任意 $n\in\mathbb N^{*}$，$\dfrac{a_{n+1}-a_{1}}{n}=a_{2}-a_{1}$，所以 $a_{n+1}=n(a_{2}-a_{1})+a_{1}$．于是数列 $\{a_{n}\}$ 为等差数列．
综上（i）（ii），原命题得证．
设数列 $\left\{ {a_n}\right\}$ 的公差为 $d$，若 $d = 0$，则所述等式显然成立；
若 $d \ne 0$，则\[\dfrac{1}{a_{k}a_{k+1}}=\dfrac{1}{d}\cdot \dfrac{a_{k+1}-a_{k}}{a_{k}a_{k+1}}=-\dfrac{1}{d}\left(\dfrac{1}{a_{k+1}}-\dfrac{1}{a_{k}}\right),\]所以\[\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}}= -\dfrac{1}{d}\left( \dfrac{1}{{{a_{n + 1}}}}-\dfrac{1}{a_1}\right) = \dfrac{1}{d} \cdot\dfrac{{{a_{n + 1}} - {a_1}}}{{{a_1}{a_{n + 1}}}}= \dfrac{n}{{{a_1}{a_{n + 1}}}}.\]（ii）充分性
依题意有\[\begin{split}&\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}} = \dfrac{n}{{{a_1}{a_{n + 1}}}}, \\& \dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + \cdots + \dfrac{1}{{{a_n}{a_{n + 1}}}} + \dfrac{1}{{{a_{n + 1}}{a_{n + 2}}}} = \dfrac{n + 1}{{{a_1}{a_{n + 2}}}}. \end{split}\]两式相减得\[\dfrac{1}{{{a_{n + 1}}{a_{n + 2}}}} = \dfrac{n + 1}{{{a_1}{a_{n + 2}}}} - \dfrac{n}{{{a_1}{a_{n + 1}}}},\]即\[{a_1} = \left(n + 1\right){a_{n + 1}} - n{a_{n + 2}}\]从而可得\[{a_1} = n{a_n} - \left(n - 1\right){a_{n + 1}}, \]因此对任意 $n\in\mathbb N^{*}$，$\dfrac{a_{n+1}-a_{1}}{n}=a_{2}-a_{1}$，所以 $a_{n+1}=n(a_{2}-a_{1})+a_{1}$．于是数列 $\{a_{n}\}$ 为等差数列．
综上（i）（ii），原命题得证．
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