2025高考数学一轮复习-6.4-数列求和-专项训练【含答案】

这是一份2025高考数学一轮复习-6.4-数列求和-专项训练【含答案】，共7页。
1.数列112,314,518,7116,…,(2n-1)+12n,…的前n项和Sn等于( )
A.n2+1-12n B.2n2-n+1-12n
C.n2+1-12n-1D.n2-n+1-12n
2.数列{an}的前n项和为Sn,已知Sn=1-2+3-4+…+(-1)n-1·n,则S17等于( )
A.9B.8C.17D.16
3.数列{an}的通项公式是an=1n+n+1,前n项和为9,则n等于( )
A.9 B.99 C.10D.100
4.已知数列{an}的前n项和Sn满足Sn=n2+n,则数列{4anan+1}的前8项的和为( )
A.67 B.78 C.89 D.910
5.已知函数f(n)=n2,n为奇数,-n2,n为偶数,且an=f(n)+f(n+1),则a1+a2+a3+…+
a100等于( )
A.0 B.100
C.-100D.10 200
6.有穷数列1,1+2,1+2+4,…,1+2+4+…+2n-1所有项的和为 .
7.已知数列{an}满足a1=1,且an+1+an=n-1 009(n∈N*),则其前2 023项之和S2 023= .
8.已知单调递增的等差数列{an}的前n项和为Sn,且S4=20,a2,a4,a8成等比数列.
(1)求数列{an}的通项公式;
(2)若bn=2an+1-3n+2,求数列{bn}的前n项和Tn.
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9.若数列{an}满足an=(-1)n-1(12n-1+12n+1),Sn为其前n项和,则下列命题正确的是( )
A.Sn1
C.Sn有最小值
D.Sn无最大值
10.数列22+122-1,32+132-1,…,(n+1)2+1(n+1)2-1的前10项和为( )
A.1755 B.111112
C.1143132D.1189132
11.已知数列{nan}的前n项和为Sn,且an=2n,则使得Sn-nan+1+500,其前n项和为Sn,若S3=6,且a1,a2,1+a3成等比数列.
(1)求数列{an}的通项公式;
(2)若bn=an+2-an,求数列{bn}的前n项和Tn.
13.已知等差数列{an}中,a3=3,a6=6,且bn=an+1,n为奇数,2an,n为偶数.
(1)求数列{bn}的通项公式及前20项和;
(2)若cn=b2n-1b2n,记数列{cn}的前n项和为Sn,求Sn.
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14.已知数列{an}的前n项和为Sn,满足S3=3a2+2,且2an=Sn+a1.
(1)求{an}的通项公式;
(2)数列{bn}满足1b1+2b2+3b3+…+nbn=an+1-2,求{bn}的前n项和Tn.
参考答案
【A级 基础巩固】
1.解析:该数列的通项公式为an=(2n-1)+12n,
则Sn=[1+3+5+…+(2n-1)]+(12+122+…+12n)=n2+1-12n.故选A.
2.解析:S17=1-2+3-4+5-6+…+15-16+17=1+(-2+3)+(-4+5)+(-6+7)+…+
(-14+15)+(-16+17)=1+1+1+…+1=9.故选A.
3.解析:因为an=1n+n+1=n+1-n,
所以Sn=a1+a2+…+an=(2-1)+(3-2)+…+(n-n-1)+
(n+1-n)=n+1-1,令n+1-1=9,得n=99.故选B.
4.解析:当n≥2时,an=Sn-Sn-1=2n,
当n=1时,a1=2也符合上式,
所以an=2n(n∈N*),
所以4anan+1=42n(2n+2)=1n(n+1)=1n-1n+1,所以数列{4anan+1}的前8项的和为(11-12)+(12-13)+…+(18-19)=89.故选C.
5.解析:由题意,得a1+a2+a3+…+a100
=12-22-22+32+32-42-42+52+…+992-1002-1002+1012
=-(1+2)+(3+2)-(4+3)+…-(99+100)+(101+100)
=-(1+2+…+99+100)+(2+3+…+100+101)
=-50×101+50×103=100.故选B.
6.解析:由题意知所求数列的通项为1-2n1-2=2n-1,故由分组求和法及等比数列的求和公式可得和为2(1-2n)1-2-n=2n+1-2-n.
答案:2n+1-2-n
7.解析:S2 023=a1+(a2+a3)+(a4+a5)+…+(a2 022+a2 023),
又an+1+an=n-1 009(n∈N*), 且a1=1,
所以S2 023=1+(2-1 009)+(4-1 009)+…+(2 022-1 009)
=1+(2+4+6+…+2 022)-1 009×1 011
=1+2+2 0222×1 011-1 009×1 011=3 034.
答案:3 034
8.解:(1)设数列{an}的公差为d(d>0),
由题意得S4=20,a42=a2·a8,
即4a1+4×32d=20,(a1+3d)2=(a1+d)(a1+7d),
解得a1=2,d=2,
所以an=2+(n-1)·2=2n.
(2)由(1)得,an=2n,
所以bn=4(n+1)-3n+2,
所以Tn=4×2-33+4×3-34+…+4(n+1)-3n+2=4[2+3+…+(n+1)]-(33+
34+…+3n+2)=4n·2+n+12-27(1-3n)1-3=2n2+6n+272-3n+32.
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9.解析:数列{an}满足an=(-1)n-1(12n-1+12n+1),
当n为偶数时,Sn=1+13-13-15+15+17-…-(12n-1+12n+1)=1-12n+11.
故当n=2时,Sn最小为45;当n=1时,Sn最大为43.故选C.
10.解析:因为(n+1)2+1(n+1)2-1=1+2n(n+2)=1+(1n-1n+2),所以其前10项和为S10=10+
(1-13)+(12-14)+(13-15)+…+(19-111)+(110-112)=10+(1+12-111-112)=1143132.故选C.
11.解析:Sn=1×21+2×22+…+n·2n,
则2Sn=1×22+2×23+…+n·2n+1,
两式相减得-Sn=2+22+…+2n-n·2n+1=2(1-2n)1-2-n·2n+1,
故Sn=2+(n-1)·2n+1.
又an=2n,
所以Sn-nan+1+50=2+(n-1)·2n+1-n·2n+1+50=52-2n+1,
依题意52-2n+10,所以d=1,a1=1.所以数列{an}的通项公式an=
1+(n-1)=n,即数列{an}的通项公式an=n.
(2)bn=an+2-an=n+2-n=n+(12)n,Tn=b1+b2+…+bn=1+12+2+(12)2+…+n+(12)n
=(1+2+…+n)+[12+(12)2+(12)3+…+(12)n]=n(n+1)2+12[1-(12) n]1-12
=n(n+1)2+1-(12)n.
13.解:(1)设等差数列{an}的公差为d,则d=a6-a36-3=1,所以an=a3+(n-3)d=n,从而bn=n+1,n为奇数,2n,n为偶数,故b1+b2+b3+…+b19+b20=(2+4+…+20)+
(22+24+…+220)=10×(2+20)2+4×(1-410)1-4=110+43(410-1).
(2)因为cn=b2n-1b2n=2n·22n=2n·4n,所以Sn=2×41+4×42+6×43+…+
2n·4n,4Sn=2×42+4×43+6×44+…+2(n-1)·4n+2n·4n+1,相减得,-3Sn=
2×41+2×42+2×43+…+2×4n-2n·4n+1,所以-3Sn=8(1-4n)1-4-2n·4n+1=
(23-2n)4n+1-83,
即Sn=(23n-29)4n+1+89.
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14.解:(1)由2an=Sn+a1得Sn=2an-a1,当n≥2时,Sn-1=2an-1-a1,故Sn-Sn-1=
2(an-an-1),则an=2an-2an-1,即an=2an-1,所以anan-1=2,所以{an}是以2为公比的等比数列,
由S3=3a2+2得a1+a3=2a2+2,即a1+a1q2=2a1q+2,所以a1=2,故an=a1qn-1=
2·2n-1=2n.
(2)1b1+2b2+3b3+…+nbn=an+1-2=2n+1-2,则n≥2时,1b1+2b2+3b3+…+n-1bn-1=2n-2,两式相减得nbn=2n,故bn=n2n,又1b1=a2-2=2,则b1=12,符合bn=n2n,所以bn=n2n,
n∈N*,Tn=b1+b2+…+bn=121+222+…+n2n,则2Tn=1+221+322+…+n2n-1,则2Tn-Tn=
Tn=120+121+122+…+12n-1-n2n=1-12n1-12-n2n=2-n+22n,所以Tn=2-n+22n.