有 $n$ 支队伍参加单循环比赛，设每支队伍获胜的场数分别为 $x_i$（$i = 1 , 2 , 3 , \cdots , n$），失败的场数分别为 ${y_i}$（$i = 1 , 2 , 3 , \cdots , n$），若 $\displaystyle \sum\limits_{i = 1}^n {x_i^3} = \sum\limits_{i = 1}^n {y_i^3} $，求证：$\displaystyle \sum\limits_{i = 1}^n {x_i^4} = \sum\limits_{i = 1}^n {y_i^4} $．
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【出处】
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【答案】
略
【解析】
由单循环赛得，$x_i + {y_i} = n - 1$（$i = 1 , 2 , 3 , \cdots , n$），$\displaystyle \sum\limits_{k = 1}^n {x_i} = \sum\limits_{k = 1}^n {{y_i}} = {\rm{C}}_n^2 = \dfrac{{n\left( {n - 1} \right)}}{2}$．令 $n - 1 = k$，则 ${y_i} = k - x_i$（$i = 1 , 2 , 3 , \cdots , n$），此时$$\sum\limits_{i = 1}^n {\left( {x_i^3 - {y_i}^3} \right)} = \sum\limits_{i = 1}^n {\left[ {x_i^3 - {{\left( {k - x_i} \right)}^3}} \right]} = \sum\limits_{i = 1}^n {\left( {2x_i^3 - 3kx_i^2 + 3{k^2}x_i - {k^3}} \right)},$$于是\[\begin{split}\sum\limits_{i = 1}^n {\left( {x_i^4 - y_i^4} \right)} &= \sum\limits_{i = 1}^n {\left[ {x_i^4 - {{\left( {k - x_i} \right)}^4}} \right]} = \sum\limits_{i = 1}^n {\left( {4kx_i^3 - 6{k^2}x_i^2 + 4{k^3}x_i - {k^4}} \right)} \\&= \sum\limits_{i = 1}^n {\left[ {\left( {6{k^2}x_i^2 - 6{k^3}x_i + 2{k^4}} \right) - 6{k^2}x_i^2 + 4{k^3}x_i - {k^4}} \right]}\\& = \sum\limits_{i = 1}^n {\left( {{k^4} - 2{k^3}x_i} \right)} = {k^3}\sum\limits_{i = 1}^n {\left( {k - 2x_i} \right)} = 0.\end{split}\]从而原命题得证．
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