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】
2011年高考江苏卷
【标注】
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设 $M = \left\{ 1 \right\},{a_2} = 2 $，求 ${a_5}$ 的值；标注答案$8$解析根据题意，有$$\forall n > 1,{S_{n + 1}} + {S_{n - 1}} = 2\left( {{S_n} + {S_1}} \right) ,$$所以\[\forall n\in\mathbb N^{\ast},S_{n + 2} + S_n = 2\left( S_{n + 1} + S_1 \right),\]从而\[\forall n\in\mathbb N^{\ast},S_{n+3}+S_{n+1}=2(S_{n+2}+S_1),\]两式相减，可得\[\forall n\in\mathbb N^{\ast},a_{n+3}+a_{n+1}=2a_{n+2},\]于是数列 $\{a_n\}$ 从第 $2$ 项起为等差数列．又\[\begin{cases} a_1=1,\\ a_2=2,\\ S_3+S_1=2(S_2+S_1),\end{cases}\]解得 $a_3=4$，从而\[a_5=a_3+2(a_3-a_2)=8.\]
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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$a_n=2n-1,n\in \mathbb N^{\ast}$解析根据题意，有\[\begin{split} \forall n>3,S_{n+3}+S_{n-3}=2(S_n+S_3),\\
\forall n>4,S_{n+4}+S_{n-4}=2(S_n+S_4),\end{split}\]于是对任意 $n\in\mathbb N^{\ast}$，均有\[\begin{split} S_{n+6}+S_n=2(S_{n+3}+S_3),\\
S_{n+8}+S_n=2(S_{n+4}+S_4),\end{split}\]因此有\[\begin{cases}S_7+S_1=2(S_4+S_3),\\
S_9+S_1=2(S_5+S_4),\end{cases}\]以及\[\begin{split}a_{n+7}+a_{n+1}=2a_{n+4},\\
a_{n+9}+a_{n+1}=2a_{n+5},\end{split}\]于是数列 $\{a_n\}$ 从第 $2$ 项起每隔 $3$ 项以及每隔 $4$ 项均构成等差数列．设 $d_1,d_2,d_3,d_4,d_5,d_6,d_7$ 分别为等差数列\[\begin{split}A_1&:a_2,a_5,a_8,a_{11},a_{14},a_{17},a_{20},\cdots,\\
A_2&:a_3,a_6,a_9,a_{12},a_{15},a_{18},\cdots,\\
A_3&:a_4,a_7,a_{10},a_{13},a_{16},a_{19}\cdots,\\
A_4&:a_2,a_6,a_{10},a_{14},a_{18},\cdots,\\
A_5&:a_3,a_7,a_{11},a_{15},a_{19},\cdots,\\
A_6&:a_4,a_8,a_{12},a_{16},\cdots,\\
A_7&:a_5,a_9,a_{13},a_{17},\cdots,\end{split}\]的公差，于是\[\begin{cases} a_{14}-a_2=4d_1=3d_4,\\
a_{15}-a_3=4d_2=3d_5,\\
a_{16}-a_4=4d_3=3d_6,\\
a_{17}-a_5=4d_1=3d_7,\\
a_{18}-a_6=4d_2=3d_4,\\
a_{19}-a_7=4d_3=3d_5,\\
a_{20}-a_8=4d_1=3d_6
,\end{cases}\]因此可得\[4d_1=4d_2=4d_3=3d_4=3d_5=3d_6=3d_7,\]不妨设\[\begin{cases}
d_1=d_2=d_3=3d,\\
d_4=d_5=d_6=d_7=4d,\end{cases}\]进而当 $n\geqslant 2$ 时，有\[a_{n+1}-a_n=(a_{n+9}-8d)-a_n=(a_{n+9}-a_n)-8d=9d-8d=d,\]因此数列 $\{a_n\}$ 从第 $2$ 项起是公差为 $d$ 的等差数列．此时有\[\begin{cases}S_3=a_1+2a_2+d,\\
S_4=a_1+3a_2+3d,\\
S_5=a_1+4a_2+6d,\\
S_7=a_1+6a_2+15d,\\
S_9=a_1+8a_2+28d,\end{cases}\]又 $a_1=1$，代入\[\begin{cases}S_7+S_1=2(S_4+S_3),\\
S_9+S_1=2(S_5+S_4),\end{cases}\]解得\[\begin{cases} a_2=3,\\ d=2,\end{cases}\]于是可得 $a_n=2n-1,n\in\mathbb N^{\ast}$．
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