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2022年高考数学上海21

  2022-12-16 20:34:06  

(18分)数列$\{a_{n}\}$对任意$n\in N^{*}$且$n\geqslant 2$,均存在正整数$i\in [1$,$n-1]$,满足$a_{n+1}=2a_{n}-a_{i}$,$a_{1}=1$,$a_{2}=3$.
(1)求$a_{4}$可能值;
(2)命题$p$:若$a_{1}$,$a_{2}$,$\dotsb$,$a_{8}$成等差数列,则$a_{9} < 30$,证明$p$为真,同时写出$p$逆命题$q$,并判断命题$q$是真是假,说明理由;
(3)若$a_{2m}=3^{m}$,$(m\in N^{*})$成立,求数列$\{a_{n}\}$的通项公式.
分析:(1)利用递推关系式可得$a_{3}=5$,然后计算$a_{4}$的值即可;
(2)由题意可得$a_{n}=2{n}-1({n}\in [1,8],{n}\in {N}^{*})$,则$a_{9}=2a_{8}-a_{i} < 30$,从而命题为真命题,给出反例可得命题$q$为假命题.
(3)由题意可得$a_{2m+2}=2a_{2m+1}-a_{i}(i\leqslant 2m)$,$a_{2m+1}=2a_{2m}-a_{j}(j\leqslant 2m-1)$,然后利用数学归纳法证明数列单调递增,最后分类讨论即可确定数列的通项公式.
解:(1)$a_{3}=2a_{2}-a_{1}=5$,$a_{4}=2a_{3}-a_{2}=7$或$a_{4}=2a_{3}-a_{1}=9$.
(2)$\because a_{1}$,$a_{2}$,$a_{3}$,$a_{4}$,$a_{5}$,$a_{6}$,$a_{7}$,$a_{8}$为等差数列,$\therefore$${d}=2,a_{n}=2{n}-1({n}\in [1,8],{n}\in {N}^{*})$,
$a_{9}=2a_{8}-a_{i}=30-a_{i} < 30$.
逆命题$q$:若$a_{9} < 30$,则$a_{1}$,$a_{2}$,$a_{3}$,$a_{4}$,$a_{5}$,$a_{6}$,$a_{7}$,$a_{8}$为等差数列是假命题,举例:
$a_{1}=1$,$a_{2}=3$,$a_{3}=5$,$a_{4}=7$,$a_{5}=9$,$a_{6}=11$,$a_{7}=13$,$a_{8}=2a_{7}-a_{5}=17$,$a_{9}=2a_{8}-a_{7}=21$.
(3)因为$a_{2m}=3^{m}$,
$\therefore$$a_{2m+2}=3^{m+1},a_{2m+2}=2a_{2m+1}-a_{i}(i\leqslant 2m)$,$a_{2m+1}=2a_{2m}-a_{j}(j\leqslant 2m-1)$,
$\therefore a_{2m+2}=4a_{2m}-2a_{j}-a_{i}$,
$\therefore$$2a_{j}+a_{i}=4a_{2m}-a_{2m+2}=4\times 3^{m}-3^{m+1}=3^{m}=a_{2m}$,
以下用数学归纳法证明数列单调递增,即证明$a_{n+1} > a_{n}$恒成立:
当$n=1$,$a_{2} > a_{1}$明显成立,
假设$n=k$时命题成立,即$a_{k} > a_{k-1} > a_{k-1}\dotsb  >  > a_{2} > a_{1} > 0$,
则$a_{k+1}-a_{k}=2a_{k}-a_{i}-a_{k}=a_{k}-a_{i} > 0$,则$a_{k+1} > a_{k}$,命题得证.
回到原题,分类讨论求解数列的通项公式:
1.若$j=2$$m-1$,则$a_{2m}=2a_{j}+a_{i}=2a_{2m-1}+a_{i} > a_{2m-1}-a_{i}$矛盾,
2.若$j=2$$m-2$,则$a_{j}=3^{m-1}$,$\therefore$$a_{i}=3^{m}-2a_{j}=3^{m-1}$,$\therefore i=2m-2$,
此时$a_{2m+1}=2a_{2m}-a_{j}=2\times 3^{m}-3^{m-1}=5\times 3^{m-1}$,
$\therefore$$a_{n}=\left\{\begin{array}{cc}{1}&{n=1}\\ {5\times 3^{\frac{n-3}{2}}}&{n=2k+1,k\in N^{*}}\\ {3^{\frac{n}{2}}}&{n=2k,k\in N^{*}}\end{array}\right.$,
3.若$j < 2$$m-2$,则$2a_{j} < 2\times 3^{m-1}$,
$\therefore$$a_{i}=3^{m}-2a_{j} > 3^{m-1}$,$\therefore j=2m-1$,
$\therefore a_{2m+2}=2a_{2m+1}-a_{2m-1}$(由(2)知对任意$m$成立),
$a_{6}=2a_{5}-a_{3}$,
事实上:$a_{6}=2a_{5}-a_{2}$矛盾.
综上可得$a_{n}=\left\{\begin{array}{cc}{1}&{n=1}\\ {5\times 3^{\frac{n-3}{2}}}&{n=2k+1,k\in N^{*}}\\ {3^{\frac{n}{2}}}&{n=2k,k\in N^{*}}\end{array}\right.$.
点评:本题主要考查数列中的递推关系式,数列中的推理问题,数列通项公式的求解等知识,属于难题.

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