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2024年高考数学第一轮复习讲义第六章6.5 数列求和(学生版+解析)
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这是一份2024年高考数学第一轮复习讲义第六章6.5 数列求和(学生版+解析),共20页。
知识梳理
数列求和的几种常用方法
1.公式法
直接利用等差数列、等比数列的前n项和公式求和.
(1)等差数列的前n项和公式:
Sn=____________=__________________.
(2)等比数列的前n项和公式:
Sn=eq \b\lc\{\rc\ (\a\vs4\al\c1( ,q=1,, = ,q≠1.))
2.分组求和法与并项求和法
(1)分组求和法
若一个数列是由若干个等差数列或等比数列或可求和的数列组成,则求和时可用分组求和法,分别求和后相加减.
(2)并项求和法
一个数列的前n项和中,可两两结合求解,则称之为并项求和.形如an=(-1)nf(n)类型,可采用两项合并求解.
3.错位相减法
如果一个数列的各项是由一个等差数列和一个等比数列的对应项之积构成的,那么这个数列的前n项和即可用此法来求,如等比数列的前n项和公式就是用此法推导的.
4.裂项相消法
把数列的通项拆成两项之差,在求和时中间的一些项可以相互抵消,从而求得其和.
常见的裂项技巧
(1)eq \f(1,nn+1)=eq \f(1,n)-eq \f(1,n+1).
(2)eq \f(1,nn+2)=eq \f(1,2)eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,n)-\f(1,n+2))).
(3)eq \f(1,2n-12n+1)=eq \f(1,2)eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,2n-1)-\f(1,2n+1))).
(4)eq \f(1,\r(n)+\r(n+1))=eq \r(n+1)-eq \r(n).
(5)eq \f(1,nn+1n+2)=eq \f(1,2)eq \b\lc\[\rc\](\a\vs4\al\c1(\f(1,nn+1)-\f(1,n+1n+2))).
常用结论
常用求和公式
(1)1+2+3+4+…+n=eq \f(nn+1,2).
(2)1+3+5+7+…+(2n-1)=n2.
(3)12+22+32+…+n2=eq \f(nn+12n+1,6).
(4)13+23+33+…+n3=eq \b\lc\[\rc\](\a\vs4\al\c1(\f(nn+1,2)))2.
思考辨析
判断下列结论是否正确(请在括号中打“√”或“×”)
(1)如果数列{an}为等比数列,且公比不等于1,则其前n项和Sn=eq \f(a1-an+1,1-q).( )
(2)求Sn=a+2a2+3a3+…+nan时,只要把上式等号两边同时乘a即可根据错位相减法求得.( )
(3)已知等差数列{an}的公差为d,则有eq \f(1,anan+1)=eq \f(1,d)eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,an)-\f(1,an+1))). ( )
(4)sin21°+sin22°+sin23°+…+sin288°+sin289°=44.5.( )
教材改编题
1.已知函数f(n)=eq \b\lc\{\rc\ (\a\vs4\al\c1(n2当n为奇数时,,-n2当n为偶数时,))且an=f(n)+f(n+1),则a1+a2+a3+…+a100等于( )
A.0 B.100 C.-100 D.10 200
2.数列{an}的前n项和为Sn.若an=eq \f(1,nn+1),则S5等于( )
A.1 B.eq \f(5,6) C.eq \f(1,6) D.eq \f(1,30)
3.Sn=eq \f(1,2)+eq \f(1,2)+eq \f(3,8)+…+eq \f(n,2n)等于( )
A.eq \f(2n-n-1,2n) B.eq \f(2n+1-n-2,2n)
C.eq \f(2n-n+1,2n) D.eq \f(2n+1-n+2,2n)
题型一 分组求和与并项求和
例1 (2023·菏泽模拟)已知数列{an}中,a1=1,它的前n项和Sn满足2Sn+an+1=2n+1-1.
(1)证明:数列eq \b\lc\{\rc\}(\a\vs4\al\c1(an-\f(2n,3)))为等比数列;
(2)求S1+S2+S3+…+S2n.
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延伸探究 在本例(2)中,如何求S1+S2+S3+…+Sn?
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思维升华 (1)若数列{cn}的通项公式为cn=an±bn,且{an},{bn}为等差或等比数列,可采用分组求和法求数列{cn}的前n项和.
(2)若数列{cn}的通项公式为cn=eq \b\lc\{\rc\ (\a\vs4\al\c1(an,n为奇数,,bn,n为偶数,))其中数列{an},{bn}是等比数列或等差数列,可采用分组求和法求{cn}的前n项和.
跟踪训练1 记数列{an}的前n项和为Sn,已知Sn=2an-2n+1.
(1)求数列{an}的通项公式;
(2)记bn=(-1)n·lg2eq \b\lc\[\rc\](\a\vs4\al\c1(\f(2,3)an+4-\f(4,3))),求数列{bn}的前n项和Tn.
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题型二 错位相减法求和
例2 (12分)(2021·全国乙卷)设{an}是首项为1的等比数列,数列{bn}满足bn=eq \f(nan,3).已知a1,3a2,9a3成等差数列.
(1)求{an}和{bn}的通项公式; [切入点:设基本量q]
(2)记Sn和Tn分别为{an}和{bn}的前n项和.证明:Tn
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