1、第2讲数列通项与求和一、选择题1.(2019山西太原模拟(一),9)已知数列an的前n项和Sn满足Sn+an=2n(nN*),则a7=()A.73 B.12764C.32132D.38564答案B当n2时,Sn-1+an-1=2n-2,又Sn+an=2n,所以2an-an-1=2,所以2(an-2)=an-1-2,故an-2是首项为a1-2,公比为12的等比数列,又S1+a1=2,故a1=1,所以an=-12n-1+2,故a7=2-164=12764,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.3
2、B.2C.1 D.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.(2017广东深圳一模,4)已知等比数列an的前n项和Sn=a3n-1+b,则ab=()A.-3B.-1C.1 D.3答案A等比数列an的前n项和Sn=a3n-1+b,a1=S1=a+b,a2=S2-S1=3a+b-a-b=2a,a3=S3-S2=9a+b-3a-b=6a,等比数列an中,a22=a1a3,(
3、2a)2=(a+b)6a,解得ab=-3.故选A.4.数列an满足an+1+(-1)nan=2n-1,则an的前60项和为()A.3 690 B.3 660C.1 845 D.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,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
4、 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=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a
5、n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
6、-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=11 009.8.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=