求证：$\displaystyle \sum\limits_{i = 0}^{\left[ {\frac{n}{3}} \right]} {\left[ {\dfrac{{n - 3i}}{2}} \right]} = \left[ {\dfrac{{{n^2} + 2n + 4}}{{12}}} \right]$．
【难度】
【出处】
2013年清华大学保送生试题
【标注】
【答案】
略
【解析】
用数学归纳法证明．
当 $n = 1, 2, 3, 4, 5, 6$ 时，命题显然成立．
假设当 $n = k$ 时命题成立，即$$\sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right]} {\left[ {\dfrac{{k - 3i}}{2}} \right]} = \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right],$$则当 $n=k+6$ 时，\[\begin{split}&\quad\sum\limits_{i = 0}^{\left[ {\frac{{k + 6}}{3}} \right]} {\left[ {\dfrac{{k + 6 - 3i}}{2}} \right]} - \left[ {\dfrac{{{{\left( {k + 6} \right)}^2} + 2\left( {k + 6} \right) + 4}}{{12}}} \right]\\&=\sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left( {\left[ {\dfrac{{k - 3i}}{2}} \right] + 3} \right)} - \left[ {\dfrac{{{k^2} + 2k + 4 + 12k + 48}}{{12}}} \right]\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right]} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\&= \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left( {k + 4} \right)\\&= \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left[ {\dfrac{k}{3}} \right] - k + 5\end{split}\]下面证明$$\forall k \in \mathbb N, \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\dfrac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\dfrac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left[ {\dfrac{k}{3}} \right] - k + 5 = 0,$$当 $k = 6m$ 时，$$ LHS= \left[ {\dfrac{{6m - 3\left( {2m + 1} \right)}}{2}} \right] + \left[ {\dfrac{{6m - 3\left( {2m + 2} \right)}}{2}} \right] + 3 \cdot 2m - 6m + 5 = 0,$$当 $k = 6m + 1$ 时，$$ RHS= \left[ {\dfrac{{6m + 1 - 3\left( {2m + 1} \right)}}{2}} \right] + \left[ {\dfrac{{6m + 1 - 3\left( {2m + 2} \right)}}{2}} \right] + 3 \cdot 2m - \left( {6m + 1} \right) + 5 = 0.$$以下略．
当 $n = 1, 2, 3, 4, 5, 6$ 时，命题显然成立．
假设当 $n = k$ 时命题成立，即$$\sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right]} {\left[ {\dfrac{{k - 3i}}{2}} \right]} = \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right],$$则当 $n=k+6$ 时，\[\begin{split}&\quad\sum\limits_{i = 0}^{\left[ {\frac{{k + 6}}{3}} \right]} {\left[ {\dfrac{{k + 6 - 3i}}{2}} \right]} - \left[ {\dfrac{{{{\left( {k + 6} \right)}^2} + 2\left( {k + 6} \right) + 4}}{{12}}} \right]\\&=\sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left( {\left[ {\dfrac{{k - 3i}}{2}} \right] + 3} \right)} - \left[ {\dfrac{{{k^2} + 2k + 4 + 12k + 48}}{{12}}} \right]\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right]} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\&= \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left( {k + 4} \right)\\&= \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left[ {\dfrac{k}{3}} \right] - k + 5\end{split}\]下面证明$$\forall k \in \mathbb N, \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\dfrac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\dfrac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left[ {\dfrac{k}{3}} \right] - k + 5 = 0,$$当 $k = 6m$ 时，$$ LHS= \left[ {\dfrac{{6m - 3\left( {2m + 1} \right)}}{2}} \right] + \left[ {\dfrac{{6m - 3\left( {2m + 2} \right)}}{2}} \right] + 3 \cdot 2m - 6m + 5 = 0,$$当 $k = 6m + 1$ 时，$$ RHS= \left[ {\dfrac{{6m + 1 - 3\left( {2m + 1} \right)}}{2}} \right] + \left[ {\dfrac{{6m + 1 - 3\left( {2m + 2} \right)}}{2}} \right] + 3 \cdot 2m - \left( {6m + 1} \right) + 5 = 0.$$以下略．
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