Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\dfrac{a}{{2016}} = \dfrac{b}{{2017}} = \dfrac{c}{{2018}}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{a}{{2016}} = \dfrac{b}{{2017}} = \dfrac{{a - b}}{{2016 - 2017}} = - \left( {a - b} \right)\\
\dfrac{b}{{2017}} = \dfrac{c}{{2018}} = \dfrac{{b - c}}{{2017 - 2018}} = - \left( {b - c} \right)\\
\dfrac{a}{{2016}} = \dfrac{c}{{2018}} = \dfrac{{c - a}}{{2018 - 2016}} = \dfrac{{c - a}}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {\dfrac{a}{{2016}}} \right)^2} = \left( { - \left( {a - b} \right)} \right)\left( { - \left( {b - c} \right)} \right) = \left( {a - b} \right)\left( {b - c} \right)\\
{\left( {\dfrac{a}{{2016}}} \right)^2} = \dfrac{{{{\left( {c - a} \right)}^2}}}{4}
\end{array} \right.\\
\Rightarrow \left( {a - b} \right)\left( {b - c} \right) = \dfrac{{{{\left( {c - a} \right)}^2}}}{4}\\
\Rightarrow 4\left( {a - b} \right)\left( {b - c} \right) = {\left( {c - a} \right)^2}
\end{array}$
