所属成套资源:2024年高考数学复习二轮讲义(考前回顾+思想方法+六专题)
专题三 微重点6 子数列与增减项问题--2024年高考数学复习二轮讲义
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这是一份专题三 微重点6 子数列与增减项问题--2024年高考数学复习二轮讲义,共4页。
考点一 奇数项、偶数项
例1 (2023·新高考全国Ⅱ)已知{an}为等差数列,bn=eq \b\lc\{\rc\ (\a\vs4\al\c1(an-6,n为奇数,,2an,n为偶数.))记Sn,Tn分别为数列{an},{bn}的前n项和,S4=32,T3=16.
(1)求{an}的通项公式;
(2)证明:当n>5时,Tn>Sn.
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规律方法 (1)数列中的奇、偶项问题的常见题型
①数列中连续两项和或积的问题(an+an+1=f(n)或an·an+1=f(n));
②含有(-1)n的类型;
③含有{a2n},{a2n-1}的类型;
④已知条件明确的奇偶项问题.
(2)对于通项公式分奇、偶不同的数列{an}求Sn时,我们可以分别求出奇数项的和与偶数项的和,也可以把a2k-1+a2k看作一项,求出S2k,再求S2k-1=S2k-a2k.
跟踪演练1 (2023·郑州模拟)已知数列{an}满足a1=3,an=an-1+2n-1(n≥2,n∈N*).
(1)求数列{an}的通项公式;
(2)令bn=an-1+(-1)nlg2(an-1),求数列{bn}的前n项和Tn.
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考点二 两数列的公共项
例2 已知数列{an}的前n项和Sn=eq \f(3n2+n,2),{bn}为等比数列,公比为2,且b1,b2+1,b3为等差数列.
(1)求{an}与{bn}的通项公式;
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(2)把数列{an}和{bn}的公共项由小到大排成的数列记为{cn},求数列{cn}的前n项和Tn.
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规律方法 两个等差数列的公共项是等差数列,且公差是两等差数列公差的最小公倍数;两个等比数列的公共项是等比数列,公比是两个等比数列公比的最小公倍数.
跟踪演练2 (2023·邵阳模拟)数列{2n-1}和数列{3n-2}的公共项从小到大构成一个新数列{an},数列{bn}满足bn=eq \f(an,2n),则数列{bn}的最大项等于________.
考点三 数列有关增减项问题
例3 已知等比数列{an}的前n项和Sn=2n+r,其中r为常数.
(1)求r的值;
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(2)设bn=2(1+lg2an),若数列{bn}中去掉数列{an}的项后余下的项按原来的顺序组成数列{cn},求c1+c2+c3+…+c100的值.
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规律方法 解决此类问题的关键是通过阅读、理解题意,要弄清楚增加了(减少了)多少项,增加(减少)的项有什么特征,在求新数列的和时,一般采用分组求和法,即把原数列部分和增加(减少)部分分别求和,再相加(相减)即可.
跟踪演练3 (2023·无锡模拟)设等比数列{an}的首项为a1=2,公比为q(q为正整数),且满足3a3是8a1与a5的等差中项;数列{bn}满足2n2-(3+bn)n+eq \f(3,2)bn=0(t∈R,n∈N*).
(1)求数列{an},{bn}的通项公式;
(2)当{bn}为等差数列时,对每个正整数k,在ak与ak+1之间插入bk个2,得到一个新数列{cn}.设Tn是数列{cn}的前n项和,试求T100.
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