2024年数学高考大一轮复习第六章 培优课 §6.7　子数列问题

§6.7　子数列问题

子数列问题包括数列中的奇偶项、公共数列以及分段数列，是近几年高考的重点和热点，一般方法是构造新数列，利用新数列的特征(等差、等比或其他特征)求解原数列．

题型一　奇数项与偶数项

例1 (2023·南通模拟)在数列{an}中，an＝

(1)求a1，a2，a3；

(2)求数列{an}的前n项和Sn.

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思维升华 解答与奇偶项有关的求和问题的关键

(1)弄清n为奇数或偶数时数列的通项公式．

(2)弄清n为奇数时数列前n项中奇数项与偶数项的个数．

跟踪训练1 (2021·新高考全国Ⅰ)已知数列{an}满足a1＝1，an＋1＝

(1)记bn＝a2n，写出b1，b2，并求数列{bn}的通项公式；

(2)求{an}的前20项和．

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题型二　两数列的公共项

例2 数列{an}与{bn}的通项公式分别为an＝4n－1，bn＝3n＋2，它们的公共项由小到大排列组成数列{cn}，求数列{cn}的通项公式．

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思维升华 解决两个等差数列的公共项问题时，有两种方法：

(1)不定方程法：列出两个项相等的不定方程，利用数论中的整除知识，求出符合条件的项，并解出相应的通项公式；

(2)周期法：即寻找下一项．通过观察找到首项后，从首项开始向后，逐项判断变化较大(如公差的绝对值大)的数列中的项是否为另一个数列中的项，并找到规律(周期)，分析相邻两项之间的关系，从而得到通项公式．

跟踪训练2 (1)已知数列{an}，{bn}的通项公式分别为an＝4n－2(1≤n≤100，n∈N*)，bn＝6n－4(n∈N*)，由这两个数列的公共项按从小到大的顺序组成一个新的数列{cn}，则数列{cn}的各项之和为(　　)

A．6 788   B．6 800

C．6 812   D．6 824

(2)我国古代数学名著《孙子算经》载有一道数学问题：“今有物不知其数，三三数之剩二，五五数之剩二，七七数之剩二，问物几何？”根据这一数学问题，所有被3除余2的自然数从小到大排列组成数列{an}，所有被5除余2的自然数从小到大排列组成数列{bn}，把{an}和{bn}的公共项从小到大排列得到数列{cn}，则(　　)

A．a3＋b5＝c3   B．b28＝c10

C．a5b2>c8   D．c9－b9＝a26

题型三　分段数列

例3 (1)记Sn为数列{an}的前n项和，Sn＝，则an＝________.

听课记录：______________________________________________________________

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(2)已知数列{an}是公差不为0的等差数列，a1＝，数列{bn}是等比数列，且b1＝a1，b2＝－a3，b3＝a4，数列{bn}的前n项和为Sn.

①求数列{bn}的通项公式；

②设cn＝求{cn}的前n项和Tn.

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思维升华 (1)利用等差数列的通项公式与等比中项性质列式可解得等差数列的公差和等比数列的公比，进而可得所求通项公式．

(2)对n分类讨论，结合等差数列与等比数列的求和公式可得所求和．

跟踪训练3 (1)已知数列{an}满足an＝若数列{an}的前n项和为Sn，则当λ＝1时，S11等于(　　)

A.   B.

C.   D.

(2)已知数列：1，1，2，1，2，4，1，2，4，8，1，2，4，8，16，…，即此数列第一项是20，接下来两项是20，21，再接下来三项是20，21，22，依此类推，设Sn是此数列的前n项和，则S2 024等于(　　)

A．264＋190   B．263＋190

C．264＋62   D．263＋62