设 $M$ 为部分正整数组成的集合，数列 $\left\{ {a_n}\right\} $ 的首项 ${a_1} = 1$，前 $n$ 项和为 ${S_n}$，已知对任意整数 $k \in M$，当整数 $n > k$ 时，${S_{n + k}} + {S_{n - k}} = 2\left({S_n} + {S_k}\right)$ 都成立．
【难度】
【出处】
无
【标注】
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设 $M = \left\{ 1 \right\},{a_2} = 2 $，求 ${a_5}$ 的值；标注答案解析$\because$ $k = 1 $，$\therefore$ $\forall n > 1,$ ${S_{n + 1}} + {S_{n - 1}} = 2\left( {{S_n} + {S_1}} \right) $，所以\[{S_{n + 2}} + {S_n} = 2\left( {{S_{n + 1}} + {S_1}} \right),\]即\[{a_{n + 2}} + {a_n} = 2{a_{n + 1}},\]所以，$n > 1$ 时，$\left\{ {a_n} \right\}$ 成等差数列，而\[{a_2} = 2, {S_2} = 3,{S_3} = 2\left( {{S_2} + {S_1}} \right) - {S_1} = 7 ,\]$\therefore$ ${a_3} = 4 $，$\therefore$ ${a_5} = 8$．
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设 $M = \left\{ {3,4} \right\}$，求数列 $\left\{ {a_n}\right\} $ 的通项公式．标注答案解析由题意：\[\begin{split} & \forall n > 3,{S_{n + 3}} + {S_{n - 3}} = 2\left( {{S_n} + {S_3}} \right) ,\quad\cdots\cdots ① \\ & \forall n > 4,{S_{n + 4}} + {S_{n - 4}} = 2\left( {{S_n} + {S_4}} \right), \quad\cdots\cdots ② \\ & \forall n > 3,{S_{n + 4}} + {S_{n - 2}} = 2\left({S_{n + 1}} + {S_3}\right) ,\quad\cdots\cdots ③ \\ & \forall n > 4,{S_{n + 5}} + {S_{n - 3}} = 2\left({S_{n + 1}} + {S_4}\right), \quad\cdots\cdots ④ \end{split}\]当 $n \geqslant 5$ 时，由 ①② 得：\[{a_{n + 4}} - {a_{n - 3}} = 2{a_4},\quad\cdots\cdots ⑤ \]由 ③④ 得：\[{a_{n + 5}} - {a_{n - 2}} = 2{a_4},\quad\cdots\cdots ⑥ \]由 ①③ 得：\[{a_{n + 4}} + {a_{n - 2}} = 2{a_{n + 1}},\quad\cdots\cdots ⑦ \]由 ②④ 得：\[{a_{n + 5}} + {a_{n - 3}} = 2{a_{n + 1}},\quad\cdots\cdots ⑧ \]由 ⑦⑧ 知：${a_{n + 4}},{a_{n + 1}},{a_{n - 2}}$ 成等差，${a_{n + 5}},{a_{n + 1}},{a_{n - 3}}$ 成等差；
设公差分别为：${d_1},{d_2} $，由 ⑤⑥ 得：\[ \begin{split} {a_{n + 5}} &= {a_{n - 3}} + 2{d_2} \\&= {a_{n + 4}} - 2{a_4} + 2{d_2},\quad\cdots\cdots ⑨ \\ {a_{n + 4}} &= {a_{n - 2}} + 2{d_1} \\&= {a_{n + 5}} - 2{a_4} + 2{d_1},\quad\cdots\cdots⑩ \end{split}\]由 ⑨ ⑩得：\[\begin{split}{a_{n + 5}} - {a_{n + 4}} & = {d_2} - {d_1}, \\ 2{a_4} & = {d_1} + {d_2}, \\ {a_{n - 2}} - {a_{n - 3}} & = {d_2} - {d_1};\end{split}\]$\therefore$ $\left\{ {a_n} \right\}\left( {n \geqslant 2} \right)$ 成等差，设公差为 $d$，在 ①② 中分别取 $n = 4,n = 5$ 得：\[\begin{cases} 2{a_1}+6{a_2} + 15d = 2\left(2{a_1} + 5{a_2} + 4d\right), \\
2{a_1} + 8{a_2} + 28d = 2\left(2{a_1} + 7{a_2} + 9d\right) ,
\end{cases}\]即\[\begin{cases} 4{a_2} - 7d = - 2, \\ 3{a_2} - 5d = - 1 , \end{cases}\]$\therefore$ ${a_2} = 3,d = 2 $，$\therefore$ ${a_n} = 2n - 1\left(n\geqslant 2\right)$，经检验，$n=1$ 时也符合．
所以 $a_n=2n-1,n\in \mathbb N^*$．
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