《2025年创新设计高考总复习高三数学人教版A版》

例 3 (2024·新高考Ⅰ卷)设$m$为正整数，数列$a_{1}$，$a_{2}$，$\cdots $，$a_{4m + 2}$是公差不为$0$的等差数列，若从中删去两项$a_{i}$和$a_{j}(i\lt j)$后剩余的$4m$项可被平均分为$m$组，且每组的$4$个数都能构成等差数列，则称数列$a_{1}$，$a_{2}$，$\cdots $，$a_{4m + 2}$是$(i,j)$－可分数列。
(1)写出所有的$(i,j)$，$1\leqslant i\lt j\leqslant 6$，使得数列$a_{1}$，$a_{2}$，$\cdots $，$a_{6}$是$(i,j)$－可分数列；
(2)当$m\geqslant 3$时，证明：数列$a_{1}$，$a_{2}$，$\cdots $，$a_{4m + 2}$是$(2,13)$－可分数列；
(3)从$1$，$2$，$\cdots $，$4m + 2$中一次任取两个数$i$和$j(i\lt j)$，记数列$a_{1}$，$a_{2}$，$\cdots $，$a_{4m + 2}$是$(i,j)$－可分数列的概率为$P_{m}$，证明：$P_{m}\gt \dfrac {1}{8}$。

答案： 1. 解
(1)满足题意的所有$(i,j)$为(1,2),(1,6),(5,6).
(2)证明 当$m = 3$时,删去$a_{2},a_{13}$,其余项可分为以下3组:$a_{1},a_{4},a_{7},a_{10}$为第1组,$a_{3},a_{6},a_{9},a_{12}$为第2组,$a_{5},a_{8},a_{11},a_{14}$为第3组,当$m＞3$时,删去$a_{2},a_{13}$,其余项可分为以下$m$组;$a_{1},a_{4},a_{7},a_{10}$为第1组,$a_{3},a_{6},a_{9},a_{12}$为第2组,$a_{5},a_{8},a_{11},a_{14}$为第3组,$a_{15},a_{16},a_{17},a_{18}$为第4组,$a_{19},a_{20},a_{21},a_{22}$为第5组,$\cdots\cdots$,$a_{4m - 1},a_{4m},a_{4m + 1},a_{4m + 2}$为第$m$组,可知每组的4个数都能构成等差数列,故数列$a_{1},a_{2},\cdots,a_{4m + 2}$是$(2,13)$—可分数列.
(3)证明 易知$a_{1},a_{2},\cdots,a_{4m + 2}$是$(i,j)$—可分数列$\Rightarrow1,2,\cdots,4m + 2$是$(4p + 1,4q + 2)$—可分数列,其中$p,q\in\{0,1,\cdots,m\}$.当$q - p＞1$时,删去$4p + 1,4q + 1$,将$1\sim4p$与$4q + 3\sim4m + 2$从小到大,每4项分为1组,可知每组的4个数成等差数列.考虑$4p + 1,4p + 3,4p + 4,\cdots,4q,4q + 2$是否可分,等同于考虑$1,3,4,\cdots,4t,4t + 2$是否可分,其中$t = q - p＞1$,可分为$(1,t + 1,2t + 1,3t + 1),(3,t + 2,2t + 3,3t + 4),\cdots,(t,2t,3t,4t + 2)$,每组4个数都能构成等差数列.故数列1,2,\cdots,4m + 2是$(4p + 2,4q + 1)$—可分数列,$p,q$且$q - p＞1$的可能取值方法数为$C_{m + 1}^{2}-m=\frac{(m - 1)m}{2}+\frac{(m + 1)m}{2}$.从而$P_{m}\geqslant\frac{m^{2}+m + 1}{8m^{2}+6m + 1}＞\frac{1}{8}$.

训练 3 (2025·广州调研)对于每项均是正整数的数列$P:a_{1}$，$a_{2}$，$\cdots $，$a_{n}$，定义变换$T_{1}$，$T_{1}$将数列$P$变换成数列$T_{1}(P):n$，$a_{1}-1$，$a_{2}-1$，$\cdots $，$a_{n}-1$。对于每项均是非负整数的数列$Q:b_{1}$，$b_{2}$，$\cdots $，$b_{m}$，定义$S(Q)=2(b_{1}+2b_{2}+\cdots +mb_{m})+b_{1}^{2}+b_{2}^{2}+\cdots +b_{m}^{2}$，定义变换$T_{2}$，$T_{2}$将数列$Q$各项从大到小排列，然后去掉所有为零的项，得到数列$T_{2}(Q)$。
(1)若数列$P_{0}$为$2$，$4$，$3$，$7$，求$S(T_{1}(P_{0}))$的值；
(2)对于每项均是正整数的有穷数列$P_{0}$，令$P_{k + 1}=T_{2}(T_{1}(P_{k}))$，$k\in \mathbf{N}$。
①探究$S(T_{1}(P_{0}))$与$S(P_{0})$的关系；
②证明：$S(P_{k + 1})\leqslant S(P_{k})$。

答案： 1.
(1)解 依题意,$P_{0}:2,4,3,7,T_{1}(P_{0}):4,1,3,2,6$,$S(T_{1}(P_{0}))=2(4 + 2×1+3×3+4×2+5×6)+4^{2}+1^{2}+3^{2}+2^{2}+6^{2}=2(4 + 2 + 9 + 8 + 30)+16 + 1 + 9 + 4 + 36=2×53 + 66=106 + 66=172$.
(2)①记$P_{0}:a_{1},a_{2},\cdots,a_{n}(a_{1},a_{2},\cdots,a_{n}\in N^{*})$,$T_{1}(P_{0}):n,a_{1}-1,a_{2}-1,\cdots,a_{n}-1$,$S(T_{1}(P_{0}))=2[n + 2(a_{1}-1)+3(a_{2}-1)+\cdots+(n + 1)(a_{n}-1)]+n^{2}+(a_{1}-1)^{2}+(a_{2}-1)^{2}+\cdots+(a_{n}-1)^{2}$,又$S(P_{0})=2(a_{1}+2a_{2}+3a_{3}+\cdots+na_{n})+a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$,所以$S(T_{1}(P_{0}))-S(P_{0})=2[n - 2 - 3-\cdots-(n + 1)]+n^{2}-2(a_{1}+a_{2}+\cdots+a_{n})+n + (a_{1}^{2}-2a_{1}+1+\cdots+a_{n}^{2}-2a_{n}+1)-(a_{1}^{2}+\cdots+a_{n}^{2})=2[n-\frac{(n + 3)n}{2}]+n^{2}-2\sum a_{i}+n - 2\sum a_{i}+n=2n - n(n + 3)+n^{2}-2\sum a_{i}+n - 2\sum a_{i}+n=0$,所以$S(T_{1}(P_{0}))=S(P_{0})$.
②证明 设$A$是每项均为非负整数的数列$a_{1},a_{2},\cdots,a_{n}$,当存在$1\leqslant i＜j\leqslant n$,使得$a_{i}\leqslant a_{j}$时,交换数列$A$的第$i$项与第$j$项得到数列$B$,则$S(B)-S(A)=2(i(a_{j}-a_{i})+j(a_{i}-a_{j}))+(a_{j}^{2}-a_{i}^{2}+a_{i}^{2}-a_{j}^{2})=2(i - j)(a_{j}-a_{i})\leqslant0$,当存在$1\leqslant m＜n$,使得$a_{m + 1}=a_{m + 2}=\cdots=a_{n}=0$时,记数列$a_{1},a_{2},\cdots,a_{m}$为$C$,则$S(C)=S(A)$,因此$S(T_{2}(A))\leqslant S(A)$,从而对于任意给定的数列$P_{0}$,由$P_{k + 1}=T_{2}(T_{1}(P_{k}))(k = 0,1,\cdots)$,由①知$S(T_{1}(P_{k}))=S(P_{k})$,所以$S(P_{k + 1})=S(T_{2}(T_{1}(P_{k})))\leqslant S(T_{1}(P_{k}))=S(P_{k})$.