2020届高考数学（文）课标版二轮习题：专题二第2讲　数列通项与求和 WORD版含解析.docx

1、第2讲数列通项与求和一、选择题1.(2019武昌调研)已知数列an的前n项和Sn=n2-1,则a1+a3+a5+a7+a9=()A.40B.44C.45D.49答案B解法一:因为Sn=n2-1,所以当n2时,an=Sn-Sn-1=n2-1-(n-1)2+1=2n-1,又a1=S1=0,所以an=0,n=1,2n-1,n2.所以a1+a3+a5+a7+a9=0+5+9+13+17=44.故选B.解法二:因为Sn=n2-1,所以当n2时,an=Sn-Sn-1=n2-1-(n-1)2+1=2n-1,又a1=S1=0,所以an=0,n=1,2n-1,n2,所以an从第二项起是等差数列,a2=3,公差d

2、=2,所以a1+a3+a5+a7+a9=0+4a6=4(26-1)=44,故选B.2.已知数列an满足an+1=an-an-1(n2,nN*),a1=1,a2=2,Sn为数列an的前n项和,则S2 018=()A.3B.2C.1D.0答案Aan+1=an-an-1(n2,nN*),a1=1,a2=2,a3=1,a4=-1,a5=-2,a6=-1,a7=1,a8=2,故数列an是周期为6的周期数列,且每连续6项的和为0,故S2 018=3360+a2 017+a2 018=a1+a2=3.故选A.3.(2019洛阳尖子生第二次联考)已知数列an的前n项和为Sn,a1=1,Sn=2an+1,则Sn

3、=()A.2n-1B.32n-1C.23n-1D.12n-1答案B解法一:Sn=2an+1=2Sn+1-2Sn3Sn=2Sn+1Sn+1Sn=32,故数列Sn为等比数列,公比是32,又S1=1,所以Sn=132n-1=32n-1.故选B.解法二:当n=1时,S1=a1=2a2,则a2=12,所以S2=1+12=32,结合选项可得只有B满足,故选B.4.数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 690B.3 660C.1 845D.1 830答案D不妨令a1=1,则a2=2,a3=a5=a7=1,a4=6,a6=10,所以当n为奇数时,an=1;当n为偶数时

4、,构成以a2=2为首项,4为公差的等差数列,所以an的前60项和为S60=30+230+30(30-1)24=1 830.5.(2018河南郑州质量预测)已知数列an的前n项和为Sn,a1=1,a2=2,且an+2-2an+1+an=0(nN*),记Tn=1S1+1S2+1Sn(nN*),则T2 018=()A.4 0342 018B.2 0172 018C.4 0362 019D.2 0182 019答案C由an+2-2an+1+an=0(nN*),可得an+2+an=2an+1,所以数列an是首项为1,公差d=a2-a1=2-1=1的等差数列,通项公式为an=a1+(n-1)d=1+n-1

5、=n,则其前n项和Sn=n(a1+an)2=n(n+1)2,所以1Sn=2n(n+1)=21n-1n+1,Tn=1S1+1S2+1Sn=211-12+12-13+1n-1n+1=21-1n+1=2nn+1,故T2 018=22 0182 018+1=4 0362 019,故选C.6.已知在数列an中,a1=-60,an+1=an+3,则|a1|+|a2|+|a3|+|a30|等于()A.445B.765C.1 080D.3 105答案Ban+1=an+3,an+1-an=3.an是以-60为首项,3为公差的等差数列.an=-60+3(n-1)=3n-63.令an0,得n21.an的前20项都为

6、负值.|a1|+|a2|+|a3|+|a30|=-(a1+a2+a20)+a21+a30=-2S20+S30.Sn=a1+an2n=-123+3n2n,|a1|+|a2|+|a3|+|a30|=765.二、填空题7.已知在数列an中,a1=1,Sn为数列an的前n项和,当n2时,有2ananSn-Sn2=1成立,则S2 017=.答案11 009解析当n2时,由2ananSn-Sn2=1得2(Sn-Sn-1)=(Sn-Sn-1)Sn-Sn2=-SnSn-1,2Sn-2Sn-1=1,又2S1=2,2Sn是以2为首项,1为公差的等差数列,2Sn=n+1,故Sn=2n+1,则S2 017=11 00

7、9.8.设数列an满足a1=1,a2=3,且2nan=(n-1)an-1+(n+1)an+1(n2),则a20的值是.答案245解析2nan=(n-1)an-1+(n+1)an+1(n2),数列nan是以1为首项,2a2-a1=5为公差的等差数列,20a20=1+519=96,a20=245.9.(2019东北四市联合体模拟(一)已知在数列an中,a1=2,an+1=(n+1)ann+2an(nN*),则k=1nkak=.答案n2-12n解析由题意可知nan+1+2anan+1=(n+1)an,两边同除以anan+1,得n+1an+1-nan=2,又1a1=12,所以nan是以12为首项,2为

8、公差的等差数列,所以k=1nkak=12n+12n(n-1)2=n2-12n.10.设数列an的前n项和为Sn,且a1=1,an+an+1=12n(n=1,2,3,),则S2n+3=.答案431-14n+2解析依题意得S2n+3=a1+(a2+a3)+(a4+a5)+(a2n+2+a2n+3)=1+14+116+14n+1=1-14n+21-14=431-14n+2.三、解答题11.(2016课标全国,17,12分)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.解析(1

9、)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+350,解得a1=2,q=2,所以an=2n.(2)由题意知,S2n+1=(2n+1)(b1+b2n+1)2=(2n+1)bn+1,又S2n+1=bnbn+1,bn+10,所以bn=2n+1.令cn=bnan,则cn=2n+12n.因此Tn=c1+c2+cn=32+522+7

10、23+2n-12n-1+2n+12n,又12Tn=322+523+724+2n-12n+2n+12n+1,两式相减得12Tn=32+12+122+12n-1-2n+12n+1,所以Tn=5-2n+52n.14.(2019天津,18,13分)设an是等差数列,bn是等比数列,公比大于0.已知a1=b1=3,b2=a3,b3=4a2+3.(1)求an和bn的通项公式;(2)设数列cn满足cn=1,n为奇数,bn2,n为偶数.求a1c1+a2c2+a2nc2n(nN*).解析(1)设等差数列an的公差为d,等比数列bn的公比为q.依题意,得3q=3+2d,3q2=15+4d,解得d=3,q=3,故a

11、n=3+3(n-1)=3n,bn=33n-1=3n.所以,an的通项公式为an=3n,bn的通项公式为bn=3n.(2)a1c1+a2c2+a2nc2n=(a1+a3+a5+a2n-1)+(a2b1+a4b2+a6b3+a2nbn)=n3+n(n-1)26+(631+1232+1833+6n3n)=3n2+6(131+232+n3n).记Tn=131+232+n3n,则3Tn=132+233+n3n+1,-得,2Tn=-3-32-33-3n+n3n+1=-3(1-3n)1-3+n3n+1=(2n-1)3n+1+32.所以,a1c1+a2c2+a2nc2n=3n2+6Tn=3n2+3(2n-1)3n+1+32=(2n-1)3n+2+6n2+92(nN*).