=
⎩k
⎨
其中k∈N*.
n1n
⎪b,n=2k,
22
(i)求数列{an(cn-1)}的通项公式;
(ii)求∑ac(n∈N).
2n
*
ii
i=1
5.(2019•新课标Ⅱ)已知{an}是各项均为正数的等比数列,a1=2,a3=2a2+16.
(1)求{an}的通项公式;
(2)设bn=log2an,求数列{bn}的前n项和.
2
6.(2018•全国)已知数列{an}的前n项和为Sn,a1=,an>0,an+1(Sn+1+Sn)=2.
(1)求Sn;
(2)求
1
S1+S2
+1
S2+S3
+⋯+
1.
Sn+Sn+1
7.(2018•北京)设{an}是等差数列,且a1=ln2,a2+a3=5ln2.
(Ⅰ)求{an}的通项公式;
(Ⅱ)求ea1+ea2+⋯+ean.
8.(2017•全国)设数列{b}的各项都为正数,且b
=bn.
n
(1)证明数列⎧1⎫为等差数列;
n+1
bn+1
⎨b⎬
⎩n⎭
(2)设b1=1,求数列{bnbn+1}的前n项和Sn.
9.(2017•新课标Ⅰ)记Sn为等比数列{an}的前n项和.已知S2=2,S3=-6.
(1)求{an}的通项公式;
(2)求Sn,并判断Sn+1,Sn,Sn+2是否成等差数列.
10.(2017•新课标Ⅲ)设数列{an}满足a1+3a2+⋯+(2n-1)an=2n.
(1)求{an}的通项公式;
(2)求数列{an}的前n项和.
2n+1
11.(2016•新课标Ⅰ)已知{a}是公差为3的等差数列,数列{b}满足b=1,b=1,
n
anbn+1+bn+1=nbn.
(Ⅰ)求{an}的通项公式;
(Ⅱ)求{bn}的前n项和.
n123
12.(2016•新课标Ⅱ)Sn为等差数列{an}的前n项和,且a1=1,S7=28,记bn=[lgan],其中[x]表示不超过x的最大整数,如[0.9]=0,[lg99]=1.
(Ⅰ)求b1,b11,b101;
(Ⅱ)求数列{bn}的前1000项和.
13.(2015•福建)等差数列{an}中,a2=4,a4+a7=15.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设b=2an-2+n,求b+b+b+⋯+b
的值.
n12310
nnn
14.(2015•山东)设数列{a}的前n项和为S,已知2S=3n+3.
(Ⅰ)求{an}的通项公式;
(Ⅱ)若数列{bn},满足anbn=log3an,求{bn}的前n项和Tn.
15.(2015•新课标Ⅰ)S为数列{a}的前n项和,已知a>0,a2+2a=4S+3
nnnnnn
(I)求{an}的通项公式:
(Ⅱ)设bn
=1
anan+1
,求数列{bn}的前n项和.
16.(2015•四川)设数列{an}(n=1,2,3,⋯)的前n项和Sn满足Sn=2an-a1,且a1,a2+1,
a3成等差数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)记数列1的前n项和为T,求使得|T-1|<1
成立的n的最小值.
a
{}n
n
n1000
17.(2015•天津)已知数列{a}满足a=qa(q为实数,且q≠1),n∈N*,a=1,a=2,
nn+2n12
且a2+a3,a3+a4,a4+a5成等差数列
(1)求q的值和{an}的通项公式;
(2)
n
n
设b=log2a2n,n∈N*,求数列{b}的前n项和.
a2n-1
n
18.(2015•陕西)设f(x)是等比数列1,x,x2,,xn的各项和,其中x>0,n∈N,
n
2.
(Ⅰ)证明:
函数F(x)=f(x)-2在(1,1)内有且仅有一个零点(记为x),且x=1+1xn+1;
nn2nn22n
(Ⅱ)设有一个与上述等比数列的首项、末项、项数分别相同的等差数列,其各项和为gn(x),
比较fn(x)和gn(x)的大小,并加以证明.
19.(2015•山东)已知数列{a}是首项为正数的等差数列,数列{1}的前n项和为n.
n
(1)求数列{an}的通项公式;
anan+1
2n+1
(2)设b=(a+1)2an,求数列{b}的前n项和T.
nnnn
20.(2014•大纲版)等差数列{an}的前n项和为Sn,已知a1=13,a2为整数,且SnS4.
(1)求{an}的通项公式;
(2)
n
n
n
设b=1,求数列{b}的前n项和T.
anan+1
21.(2014•浙江)已知等差数列{an}的公差d>0,设{an}的前n项和为Sn,a1=1,S2S3=36.
(Ⅰ)求d及Sn;
mm+1m+2m+k
(Ⅱ)求m,k(m,k∈N*)的值,使得a+a+a+⋯+a=65.
n24
22.(2014•新课标Ⅰ)已知{a}是递增的等差数列,a,a是方程x2-5x+6=0的根.
(1)求{an}的通项公式;
(2)求数列
an
{2n}
的前n项和.
23.(2014•新课标Ⅱ)已知数列{an}满足a1=1,an+1=3an+1.
(Ⅰ)证明{a+1}是等比数列,并求{a}的通项公式;
n2n
(Ⅱ)证明:
1+1
a1a2
+⋯+1
an
<3.
2
24.(2014•安徽)数列{a}满足a=1,na=(n+1)a
+n(n+1),n∈N*.
n1n+1n
n
(Ⅰ)证明:
数列{a}
n
是等差数列;
an
(Ⅱ)设b=3n,求数列{b}的前n项和S.
nnn
历年高考数学真题精选(按考点分类)
专题26数列的综合(教师版)
1.(2016•新课标Ⅱ)等差数列{an}中,a3+a4=4,a5+a7=6.
(Ⅰ)求{an}的通项公式;
(Ⅱ)设bn=[an],求数列{bn}的前10项和,其中[x]表示不超过x的最大整数,如[0.9]=0,
[2.6]=2.
解:
(Ⅰ)设等差数列{an}的公差为d,
a3+a4=4,a5+a7=6.
∴⎧2a1+5d=4,
⎨2a+10d=6
⎩1
⎧a1=1
⎨d=
解得:
⎪2,
⎪⎩5
∴an
=2n+3;
55
(Ⅱ)bn=[an],
∴b1=b2=b3=1,
b4=b5=2,
b6=b7=b8=3,
b9=b10=4.
故数列{bn}的前10项和S10=3⨯1+2⨯2+3⨯3+2⨯4=24.
2.(2013•山东)设等差数列{an}的前n项和为Sn,且S4=4S2,a2n=2an+1.
(1)求数列{an}的通项公式;
(2)设数列{b}的前n项和为T且T
+an+1=λ(λ为常数).令c=b
(n∈N*)求数列{c}
n
的前n项和Rn.
nn2n
n2nn
解:
(1)设等差数列{an}的首项为a1,公差为d,由a2n=2an+1,取n=1,得a2=2a1+1,
即a1-d+1=0①
再由S
=4S,得4a+4⨯3d=4(a+a+d),即d=2a②
4212111
联立①、②得a1=1,d=2.
所以an=a1+(n-1)d=1+2(n-1)=2n-1;
(2)把a
=2n-1代入T
+an+1=λ,得T
+2n=λ,则T
=λ-2n.
n
所以b1=T1=λ-1,
n2n
n2n
n2n
当n
2时,b=T-T=(λ-2n)-(λ-2(n-1))=n-2.
nnn-12n
2n-1
2n-1
所以b=n-2,c=b
=2n-2=n-1.
n2n-1
n2n
22n-14n-1
R=c+c
+⋯+c
=0+1+
2+⋯+n-1③
n12
n4142
4n-1
1R=
1+2+⋯+n-1④
4n42434n
1(1-1)
③-④得:
3R
=1+1
+⋯+1
-n-1=44n-1
-n-1
4n442
4n-14n
1-14n
4
所以Rn
=4(1-3n+1);
94n
所以数列{c}的前n项和R=4(1-3n+1).
nn94n
3.(2011•辽宁)已知等差数列{an}满足a2=0,a6+a8=-10.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求数列{an}的前n项和S.
2n-1n
解:
(I)设等差数列{a}的公差为d,由已知条件可得⎧a1+d=0,
n⎨2a+12d=-10
⎩1
⎨d=-1
解得:
⎧a1=1,
⎩
故数列{an}的通项公式为an=2-n;
(II)设数列{an}的前n项和为S,即S
=a+a2+⋯+an①,故S=1,
2n-1n
Sn=a1+a2+⋯+an②,
2242n
当n>1时,①-②得:
n122n-11
Sn=a
+a2-a1+⋯+an-an-1-an
2122n-12n
=1-(1+1+⋯+
1)-2-n
242n-12n
=1-(1-
1
2n-1
)-2-n=n,
2n2n
所以Sn
=n,
2n-1
综上,数列{an}的前n项和S
2n-1n
=n.
2n-1
4.(2019•天津)设{an}是等差数列,{bn}是等比数列.已知a1=4,b1=6,b2=2a2-2,
b3=2a3+4.
(Ⅰ)求{an}和{bn}的通项公式;
(Ⅱ)设数列{c}满足c=1,c
⎧⎪1,2k=
⎩k
⎨
其中k∈N*.
n1n
⎪b,n=2k,
22
(i)求数列{an(cn-1)}的通项公式;
(ii)求∑ac(n∈N).
2n
*
ii
i=1
解:
(Ⅰ)设等差数列{an}的公差为d,等比数列{bn}的公比为q,
依题意有:
⎧6q=6+2d
⎩
⎨6q2=12+4d
,解得⎧d=3,
⎨q=2
⎩
∴an=4+(n-1)⨯3=3n+1,
n
b=6⨯2n-1=3⨯2n.
(Ⅱ)(i)数列{c}满足c=1,c
⎧⎪1,2k=
⎩k
⎨
其中k∈N*.
2
n1n
⎪b,n=2k,
2
2
∴an
(cn
-1)=an
(bn
-1)=(3⨯2n+1)(3⨯2n-1)=9⨯4n-1,
22
∴数列{an(cn-1)}的通项公式为:
2nn2
a(c-1)=9⨯4n-1.
=(2n⨯4+2n(2n-1)⨯3)+∑n
(9⨯4i-1)
=2n-1
2i=1
n-14(1-4n)
(3⨯2
+5⨯2)+9⨯-n
1-4
=27⨯22n-1+5⨯2n-1-n-12.(n∈N*).
5.(2019•新课标Ⅱ)已知{an}是各项均为正数的等比数列,a1=2,a3=2a2+16.
(1)求{an}的通项公式;
(2)设bn=log2an,求数列{bn}的前n项和.解:
(1)设等比数列的公比为q,
由a=2,a=2a
+16,得2q2=4q+16,
132
即q2-2q-8=0,解得q=-2(舍)或q=4.
1
∴an
=aqn-1=2⨯4n-1=22n-1;
(2)bn
=log2an
=log22n-1=2n-1,
2
b1=1,bn+1-bn=2(n+1)-1-2n+1=2,
∴数列{bn}是以1为首项,以2为公差的等差数列,
则数列{b}的前n项和T
=n⨯1+n(n-1)⨯2=n2.
nn2
2
6.(2018•全国)已知数列{an}的前n项和为Sn,a1=,an>0,an+1(Sn+1+Sn)=2.
(1)求Sn;
(2)求
1
S1+S2
+1
S2+S3
+⋯+
1.
Sn+Sn+1
2
n
解:
(1)a1=,an>0,an+1(Sn+1+Sn)=2,可得(Sn+1-Sn)(Sn+1+Sn)=2,
S可得
2
n+1
-S2=2,
n
n
即数列{S2}为首项为2,公差为2的等差数列,可得S2=2+2(n-1)=2n,
2n
由an>0,可得Sn=;
(2)
1=
1
2n+2(n+1)
Sn+Sn+1
n+n+1
=2(1
)=2(
-n),
n+1
22
即有1
S1+S2
+1
S2+S3
+⋯+
1
Sn+Sn+1
2
=2(
2
-1+3-
2+2-
3+⋯+
n+1-n)
n+1
=2(
2
-1).
7.(2018•北京)设{an}是等差数列,且a1=ln2,a2+a3=5ln2.
(Ⅰ)求{an}的通项公式;
(Ⅱ)求ea1+ea2+⋯+ean.
解:
(Ⅰ){an}是等差数列,且a1=ln2,a2+a3=5ln2.可得:
2a1+3d=5ln2,可得d=ln2,
{an}的通项公式;an=a1+(n-1)d=nln2,
(Ⅱ)ean=eln2n=2n,
aaa
123
n2(1-2n)
n+1
∴e1+e2+⋯+en=2+2
+2+⋯+2
==2
1-2
-2.
8.(2017•全国)设数列{b}的各项都为正数,且b
=bn.
n
(1)证明数列⎧1⎫为等差数列;
n+1
bn+1
⎨b⎬
⎩n⎭
(2)设b1=1,求数列{bnbn+1}的前n项和Sn.
解:
(1)证明:
数列{b}的各项都为正数,且b
=bn,
n
两边取倒数得1=bn+1=1+1,
n+1
bn+1
bn+1bnbn
故数列⎧1⎫为等差数列,其公差为1,首项为1;
⎨b⎬b
⎩n⎭1
(2)由
(1)得,1=1,1=1+(n-1)=n,
b1
故b=1,所以bb=
bnb1
1=1-1,
nnnn+1
n(n+1)nn+1
因此S
=1-1+1-1+⋯+1-1
=n.
n223
nn+1
n+1
9.(2017•新课标Ⅰ)记Sn为等比数列{an}的前n项和.已知S2=2,S3=-6.
(1)求{an}的通项公式;
(2)求Sn,并判断Sn+1,Sn,Sn+2是否成等差数列.解:
(1)设等比数列{an}首项为a1,公比为q,
则a=S-S
=-6-2=-8,则a=a3=-8,a
=a3=-8,
332
1q2q2
2qq
由a+a=2,-8+-8=2,整理得:
q2+4q+4=0,解得:
q=-2,
12q2q
则a=-2,a=(-2)(-2)n-1=(-2)n,
1n
∴{a}的通项公式a=(-2)n;
nn
a(1-qn)-2[1-(-2)n]1
(2)由
(1)可知:
S
=1==-
[2+(-2)n+1],
n1-q1-(-2)3
则Sn+1
=-1[2+(-2)n+2],S
3
n+2
=-1[2+(-2)n+3],3
由Sn+1
+
Sn+2
=-1[2+(-2)n+2]-1[2+(-2)n+3],33
=-1[4+(-2)⨯(-2)n+1+(-2)2⨯(-2)n+1],3
=-1[4+2(-2)n+1]=2⨯[-1(2+(-2)n+1)],
33
=2Sn,
即Sn+1+Sn+2=2Sn,
∴Sn+1,Sn,Sn+2成等差数列.
10.(2017•新课标Ⅲ)设数列{an}满足a1+3a2+⋯+(2n-1)an=2n.
(1)求{an}的通项公式;
(2)求数列{an}的前n项和.
2n+1
解:
(1)数列{an}满足a1+3a2+⋯+(2n-1)an=2n.
n
2时,a1+3a2+⋯+(2n-3)an-1=2(n-1).
∴(2n-1)an
=2.∴an
=2.
2n-1
当n=1时,a1=2,上式也成立.
∴an
=2.
2n-1
(2)an=
2=1-1.
2n+1(2n-1)(2n+1)2n-12n+1
∴数列{an
}的前n项和=(1-1)+(1-1)+⋯+(1
-1)=1-1
=2n.
2n+13352n-12n+12n+12n+1
11.(2016•新课标Ⅰ)已知{a}是公差为3的等差数列,数列{b}满足b=1,b=1,
n
anbn+1+bn+1=nbn.
(Ⅰ)求{an}的通项公式;
(Ⅱ)求{bn}的前n项和.
解:
(Ⅰ)anbn+1+bn+1=nbn.当n=1时,a1b2+b2=b1.
b=1,b=1,
n123
123
∴a1=