设数列 $\left\{ {a_n} \right\}$：$1, - 2, - 2,3,3,3, - 4, - 4, - 4, - 4, \cdots $，$\underbrace {{{\left( { - 1} \right)}^{k - 1}}k, \cdots ,{{\left( { - 1} \right)}^{k - 1}}k}_{k个}$，$ \cdots $，即当 $\dfrac{{\left( {k - 1} \right)k}}{2} < n \leqslant \dfrac{{k\left( {k + 1} \right)}}{2}$（$k \in {{\mathbb{N}}^*}$）时，${a_n} = {\left( { - 1} \right)^{k - 1}}k$．记 ${S_n} = {a_1} + {a_2} + \cdots + {a_n}$（$n \in {{\mathbb{N}}^*}$）．对于 $l \in {{\mathbb{N}}^*}$，定义集合 ${P_l} = \left\{ {n\left|\right. {S_n} 是 {a_n} 的整数倍,n \in {{\mathbb{N}}^*},且 1 \leqslant n \leqslant l} \right\}$．
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【出处】
2013年高考江苏卷
【标注】
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求集合 ${P_{11}}$ 中元素的个数；标注答案$ 5$．解析分别求出 $S_1$ 到 $S_{11}$，逐一验证即可．由数列 $\left\{ {a_n} \right\}$ 的定义得\[{a_1} = 1 , {a_2} = - 2 , {a_3} = - 2 , {a_4} = 3 , {a_5} = 3 ,\\ {a_6} = 3 , {a_7} = - 4 , {a_8} = - 4 , {a_9} = - 4 , {a_{10}} = - 4 ,{a_{11}} = 5,\]根据数列的前 $n$ 项和概念算得\[{S_1} = 1,{S_2} = - 1 ,{S_3} = - 3 ,{S_4} = 0 ,{S_5} = 3 ,\\ {S_6} = 6 ,{S_7} = 2 ,{S_8} = - 2,{S_9} = - 6 ,{S_{10}} = - 10,{S_{11}} = - 5,\]从而\[{S_1} = {a_1} ,{S_4} = 0 \times {a_4},{S_5} = {a_5} ,{S_6} = 2{a_6} , {S_{11}} = - {a_{11}}.\]所以集合 ${P_{11}}$ 中元素的个数为 $ 5$．
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求集合 ${P_{2000}}$ 中元素的个数．标注答案$1008$．解析求出 $S_{2i+1}$ 的通项公式，并得出其与 $2i+1$ 的倍数关系是解决本题的关键．先用数学归纳法证明：\[{S_{i\left( {2i + 1} \right)}} = - i\left( {2i + 1} \right)\left( {i \in {{\mathbb{N}}^*}} \right)\]事实上，① 当 $i = 1$ 时，${S_{i\left( {2i + 1} \right)}} = {S_3} = - 3$，$ - i\left( {2i + 1} \right) = - 3$，故原等式成立；
② 假设 $i = m$ 时成立，即 ${S_{m\left( {2m + 1} \right)}} = - m\left( {2m + 1} \right)$，则 $i = m + 1$ 时，\[\begin{split}{S_{\left( {m + 1} \right)\left( {2m + 3} \right)}} &= {S_{m\left( {2m + 1} \right)}} + {\left( {2m + 1} \right)^2} - {\left( {2m + 2} \right)^2} \\&= - m\left( {2m + 1} \right) - 4m - 3\\ &= - \left( {2{m^2} + 5m + 3} \right) \\&= - \left( {m + 1} \right)\left( {2m + 3} \right),\end{split}\]综合 ①② 可得，${S_{i\left( {2i + 1} \right)}} = - i\left( {2i + 1} \right)$．
于是\[\begin{split}{S_{\left( {i + 1} \right)\left( {2i + 1} \right)}} &= {S_{i\left( {2i + 1} \right)}} + {\left( {2i + 1} \right)^2}\\& = - i\left( {2i + 1} \right) + {\left( {2i + 1} \right)^2}\\ &= \left( {2i + 1} \right)\left( {i + 1} \right) .\end{split}\]由上可知 ${S_{i\left( {2i + 1} \right)}}$ 是 $2i + 1$ 的倍数，而\[{a_{i\left( {2i + 1} \right) + j}} = 2i + 1\left( {j = 1,2, \cdots ,2i + 1} \right),\]所以 ${S_{i\left( {2i + 1} \right) + j}} = {S_{i\left( {2i + 1} \right)}} + j\left( {2i + 1} \right)$ 是 ${a_{i\left( {2i + 1} \right) + j}}\left( {j = 1,2, \cdots ,2i + 1} \right)$ 的倍数．
又 $ {S_{\left( {i + 1} \right)\left( {2i + 1} \right)}} = \left( {i + 1} \right)\left( {2i + 1} \right) $ 不是 $ 2i + 2 $ 的倍数，而\[ {a_{\left( {i + 1} \right)\left( {2i + 1} \right) + j}} = - \left( {2i + 2} \right)\left( {j = 1,2, \cdots ,2i + 2} \right) ,\]所以\[\begin{split} {S_{\left( {i + 1} \right)\left( {2i + 1} \right) + j}} &= {S_{\left( {i + 1} \right)\left( {2i + 1} \right)}} - j\left( {2i + 2} \right)\\& = \left( {2i + 1} \right)\left( {i + 1} \right) - j\left( {2i + 2} \right)\end{split}\]不是 ${a_{\left( {i + 1} \right)\left( {2i + 1} \right) + j}}\left( {j = 1,2, \cdots ,2i + 2} \right)$ 的倍数，
故当 $l = i\left( {2i + 1} \right)$ 时，集合 ${P_l}$ 中元素的个数为\[1 + 3 + \cdots + \left( {2i - 1} \right) \overset{\left[a\right]}= {i^2}.\]（推导中用到：[a]）
于是，当 $l = i\left( {2i + 1} \right) + j\left( {1 \leqslant j \leqslant 2i + 1} \right)$ 时，集合 ${P_l}$ 中元素的个数为 ${i^2} + j$．
又 $2000 = 31 \times \left( {2 \times 31 + 1} \right) + 47,$
故集合 ${P_{2000}}$ 中元素的个数为 ${31^2} + 47 = 1008$．
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