已知 $n\in\mathbb N^{\ast}$，求证：$\displaystyle \sum\limits_{i = 0}^{\left[ {\frac{n}{3}} \right]} {\left[ {\dfrac{{n - 3i}}{2}} \right]} = \left[ {\dfrac{{{n^2} + 2n + 4}}{{12}}} \right]$．
解析 用跳跃数学归纳法证明．当 $n = 1, 2, 3, 4, 5, 6$ 时，有\[\begin{array}{c|c|c|c|c|c|c}\hline n&1&2&3&4&5&6\\ \hline \sum\limits_{i = 0}^{\left[ {\frac{n}{3}} \right]} {\left[ {\dfrac{{n - 3i}}{2}} \right]}&0&1&1&2&3&4\\ \hline {n^2} + 2n + 4&7&12&19&28&39&52\\ \hline \end{array}\] 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，有\[\begin{split}&\quad\sum\limits_{i = 0}^{\left[ {\frac{{k + 6}}{3}} \right]} {\left[ {\dfrac{{k + 6 - 3i}}{2}} \right]} - \left[ {\dfrac{{{{\left( {k + 6} \right)}^2} + 2\left( {k + 6} \right) + 4}}{{12}}} \right]\\&=\sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left( {\left[ {\dfrac{{k - 3i}}{2}} \right] + 3} \right)} - \left[ {\dfrac{{{k^2} + 2k + 4 + 12k + 48}}{{12}}} \right]\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right] + 2} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\&= \sum\limits_{i = 0}^{\left[ {\frac{k}{3}} \right]} {\left[ {\dfrac{{k - 3i}}{2}} \right]} + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 1} \right)}}{2}} \right] + \left[ {\dfrac{{k - 3 \cdot \left( {\left[ {\frac{k}{3}} \right] + 2} \right)}}{2}} \right] + 3\left( {\left[ {\dfrac{k}{3}} \right] + 3} \right) - \left[ {\dfrac{{{k^2} + 2k + 4}}{{12}}} \right] - \left( {k + 4} \right)\\ &=\left[\dfrac{k-\left[\frac k3\right]+1}{2}\right]+\left[\dfrac{k-\left[\frac k3\right]}{2}\right]+\left[\dfrac k3\right]-k,\end{split}\]而注意到 $k-\left[\frac k3\right]+1$ 和 $k-\left[\frac k3\right]$ 是相邻的两个整数，因此\[\left[\dfrac{k-\left[\frac k3\right]+1}{2}\right]+\left[\dfrac{k-\left[\frac k3\right]}2\right]=k-\left[\frac k3\right],\]因此命题得证．