1、第一章测评(时间:120分钟满分:150分)一、选择题(本大题共12小题,每小题5分,共60分)1.若数列2n+1的第k项等于3,则第3k项等于()A.3B.5C.7D.9解析:依题意2k+1=3,所以k=4,因此第3k项即第12项等于212+1=5.答案:B2.等差数列an中,若a3+a4+a5=12,则an的前7项和S7=()A.22B.24C.26D.28解析:由等差数列的性质得a3+a4+a5=3a4=12,a4=4,则S7=7(a1+a7)2=7a4=28.答案:D3.等比数列an中,a2,a6是方程x2-34x+64=0的两根,则a4等于()A.8B.-8C.8D.以上都不对解析:
2、由已知得a2+a6=340,a2a6=640,所以a20,a60,从而a40,且a42=a2a6=64,故a4=8.答案:A4.等比数列an的前n项和为Sn,已知S3=8,S6=9,则a7+a8+a9等于()A.-18B.18C.578D.558解析:由于S3,S6-S3,S9-S6成等比数列,又S3=8,S6=9,所以8,1,a7+a8+a9成等比数列,故a7+a8+a9=18.答案:B5.若数列an的通项公式是an=(-1)n(3n-2),则a1+a2+a10=()A.15B.12C.-12D.-15解析:an=(-1)n(3n-2),a1+a2+a10=-1+4-7+10-25+28=(
3、-1+4)+(-7+10)+(-25+28)=35=15.答案:A6.定义:在数列an中,若满足an+2an+1-an+1an=d(nN+,d为常数),称an为“等差比数列”.已知在“等差比数列”an中,a1=a2=1,a3=3,则a2017a2015=()A.42 0172-1B.42 0182-1C.42 0152-1D.42 0162-1解析:因为a1=a2=1,a3=3,所以a3a2-a2a1=2,所以数列an+1an是以1为首项,2为公差的等差数列,所以an+1an=2n-1.所以a2017a2015=a2017a2016a2016a2015=(22016-1)(22015-1)=4
4、20152-1,故选C.答案:C7.已知数列an中,an=12n+14,其前n项和为Sn,则数列1Sn的前8项和为()A.5845B.11645C.910D.95解析:因为an=12n+14,所以an是等差数列.从而Sn=n34+12n+142=n(n+2)4,于是1Sn=4n(n+2)=21n-1n+2,所以前8项和T8=21-13+12-14+13-15+17-19+18-110=11645.答案:B8.在函数y=f(x)的图像上有点列(xn,yn),若数列xn是等差数列,数列yn是等比数列,则函数y=f(x)的解析:式可能为()A.f(x)=2x+1B.f(x)=4x2C.f(x)=lo
5、g3xD.f(x)=34x解析:对于函数f(x)=34x图像上的点列(xn,yn),有yn=34xn,因为xn是等差数列,所以xn+1-xn=d.因此yn+1yn=34xn+134xn=34xn+1-xn=34d,这是一个与n无关的常数,故yn是等比数列,故选D.答案:D9.122-1+132-1+142-1+1(n+1)2-1的值为()A.n+12(n+2)B.34-n+12(n+2)C.34-121n+1+1n+2D.32-1n+1+1n+2解析:1(n+1)2-1=1n2+2n=1n(n+2)=121n-1n+2,122-1+132-1+142-1+1(n+1)2-1=121-13+12
6、-14+13-15+1n-1n+2=1232-1n+1-1n+2=34-121n+1+1n+2.答案:C10.已知数列an满足a1=0,且an+1=13an-2,则an的通项公式是()A.an=13n-1B.an=13n-2C.an=13n-3D.an=13n-2-3解析:由an+1=13an-2,得an+1+3=13(an+3),所以an+3是首项为0+3=3,公比为13的等比数列,于是an+3=313n-1,故an=313n-1-3,即an=13n-2-3.答案:D11.已知等比数列an满足an0,n=1,2,且a5a2n-5=22n(n3),则当n1时,log2a1+log2a3+log
7、2a2n-1等于()A.n(2n-1)B.(n+1)2C.n2D.(n-1)2解析:设公比为q,则a5a2n-5=(a1q4)(a1q2n-6)=a12q2n-2=22n,所以a1qn-1=2n,即an=2n,所以原式=log2(a1a3a2n-1)=log221+3+2n-1=log22n2=n2.答案:C12.导学号33194031已知等差数列an的通项公式an=64-4n5,设An=|an+an+1+an+12|(nN+),则当An取最小值时,n的取值为()A.16B.14C.12D.10解析:由an=64-4n50,得n16,且a16=0,所以a16-i+a16+i=0(iN+),An
8、中共13项的和,因此取n=10,则an+an+1+an+12=0,即An=0最小,故选D.答案:D二、填空题(本大题共4小题,每小题5分,共20分)13.等比数列an的前n项和为Sn,已知S1,2S2,3S3成等差数列,则an的公比为.解析:S1,2S2,3S3成等差数列,4S2=S1+3S3,4(a1+a2)=a1+3(a1+a2+a3),a2=3a3,q=13.答案:1314.(2017江苏高考)等比数列an的各项均为实数,其前n项和为Sn.已知S3=74,S6=634,则a8=.解析:设该等比数列的公比为q,则S6-S3=634-74=14,即a4+a5+a6=14.S3=74,a1+a
9、2+a3=74.由得(a1+a2+a3)q3=14,q3=1474=8,即q=2.a1+2a1+4a1=74,a1=14,a8=a1q7=1427=32.答案:3215.(2017全国2高考)等差数列an的前n项和为Sn,a3=3,S4=10,则k=1n1Sk=.解析:设等差数列的首项为a1,公差为d,由题意可知a1+2d=3,4a1+432d=10,解得a1=1,d=1.所以Sn=na1+n(n-1)2d=n(1+n)2.所以1Sn=2n(n+1)=21n-1n+1.所以k=1n1Sk=21-12+12-13+1n-1n+1=21-1n+1=2nn+1.答案:2nn+116.导学号33194
10、032设数列an满足a1=1,a2=4,a3=9,an=an-1+an-2-an-3(n4),则a2 015=.解析:由an=an-1+an-2-an-3,得an+1=an+an-1-an-2,两式相加,得an+1=2an-1-an-3,即an+1+an-3=2an-1(n4),所以数列an的奇数项和偶数项均构成等差数列.因为a1=1,a3=9,所以奇数项的公差为8,所以a2015=1+8(1008-1)=8057.答案:8 057三、解答题(本大题共6小题,共70分)17.(本小题满分10分)已知数列an中,a1=2,且an=2an-1-n+2(n2,nN+).(1)求a2,a3,并证明an
11、-n是等比数列;(2)求an的通项公式.解(1)由已知an=2an-1-n+2(n2,nN+),得a2=4,a3=7.an-n=2an-1-2n+2=2an-1-(n-1),an-nan-1-(n-1)=2.又a1-1=1,an-n是首项为1,公比为2的等比数列.(2)由(1)知an-n=12n-1,an=2n-1+n.18.(本小题满分12分)已知等差数列an,Sn为其前n项和,a5=10,S7=56.(1)求数列an的通项公式;(2)若bn=an+(3)an,求数列bn的前n项和Tn.解(1)由S7=a1+a2+a3+a4+a5+a6+a7=7a4=56,得a4=8,所以公差d=a5-a4
12、=2,an=a5+(n-5)d=2n,即an=2n.(2)将an=2n代入得bn=2n+3n,所以Tn=(2+31)+(4+32)+(6+33)+(2n+3n)=(2+4+2n)+(3+32+3n)=n(2+2n)2+3(1-3n)1-3=n2+n+3n+1-32.19.(本小题满分12分)已知等差数列an的公差d0,它的前n项和为Sn,若S5=70,且a2,a7,a22成等比数列.(1)求数列an的通项公式;(2)若数列1Sn的前n项和为Tn,求证:16Tn38.(1)解由已知,S5=5a3,a3=14,又a2,a7,a22成等比数列,由(a1+6d)2=(a1+d)(a1+21d),且d0
13、可解得a1=32d,a1=6,d=4,故数列an的通项公式为an=4n+2,nN+.(2)证明由(1)知Sn=n(a1+an)2=2n2+4n,1Sn=12n2+4n=141n-1n+2,Tn=141-13+12-14+1n-1n+2=38-141n+1+1n+2.显然,16Tn0,对于任意的正整数都成立,Sn+1Sn,即前n项和Sn组成的新数列Sn为递增数列.21.(本小题满分12分)各项均为正数的数列an满足a1=1,an+12-an2=2(nN+).(1)求数列an的通项公式;(2)求数列an22n的前n项和Sn.解(1)因为an+12-an2=2,所以数列an2是首项为1,公差为2的等
14、差数列,所以an2=1+(n-1)2=2n-1.因为an0,所以an=2n-1(nN+).(2)由(1)知,an=2n-1,所以an22n=2n-12n.所以Sn=12+322+523+2n-32n-1+2n-12n,则12Sn=122+323+524+2n-32n+2n-12n+1,-得12Sn=12+222+223+224+22n-2n-12n+1=12+2122+123+124+12n-2n-12n+1=12+2141-12n-11-12-2n-12n+1=32-2n+32n+1.所以Sn=3-2n+32n.22.导学号33194033(本小题满分12分)(2016全国甲高考)等差数列an中,a3+a4=4,a5+a7=6.(1)求an的通项公式;(2)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(1)设数列an的公差为d,由题意有2a1+5d=4,a1+5d=3,解得a1=1,d=25.所以an的通项公式为an=2n+35.(2)由(1)知,bn=2n+35.当n=1,2,3时,12n+352,bn=1;当n=4,5时,22n+353,bn=2;当n=6,7,8时,32n+354,bn=3;当n=9,10时,42n+355,bn=4.所以数列bn的前10项和为13+22+33+42=24.