实数${a_1} , {a_2} , \cdots,{a_{2017}}$满足${a_1} + {a_2} + \cdots + {a_{2017}} = 0$，且$$\left| {{a_1} - 2{a_2}} \right| = \left| {{a_2} - 2{a_3}} \right| = \cdots = \left| {{a_{2016}} - 2{a_{2017}}} \right|= \left| {{a_{2017}} - 2{a_1}} \right|.$$求证：${a_1} = {a_2} = \cdots = {a_{2017}} = 0$．
分析与解 令${b_1} = {a_1} - 2{a_2}$，${b_2} = {a_2} - 2{a_3}$，$\cdots$，${b_{2017}} = {a_{2017}} - 2{a_1}$，则
$$\left| {{b_1}} \right| = \left| {{b_2}} \right| = \cdots = \left| {{b_{2017}}} \right|,{b_1} + {b_2} + \cdots + {b_{2017}} = 0.$$设$\left| {{b_1}} \right| = \left| {{b_2}} \right| = \cdots = \left| {{b_{2017}}} \right|= m$，则${b_1} , {b_2} , \cdots ,{b_{2017}}$或者为$m$或者为$ - m$，设其中有$x$个$m$，$(2017 - x)$个$ - m$，则$${b_1} + {b_2} + \cdots + {b_{2017}} = mx + \left( { - m} \right)\left( {2017 - x} \right) = m\left( {2x - 2017} \right).$$由于$2x - 2017 \ne 0$，因此$m = 0$．
于是$${b_1} = {b_2} = \cdots = {b_{2017}} = 0,$$进而易得$${a_1} = {a_2} = \cdots = {a_{2017}} = 0.$$