2019高考数学（文科）二轮专题攻略习题：第八讲　数列求和及简单应用 WORD版含解析.docx

1、第八讲数列求和及简单应用1.数列an的前n项和Sn=n2+3n+2,则an的通项公式为()A.an=2n-2 B.an=2n+2C.an=6,n=12n-2,n2D.an=6,n=12n+2,n22.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.03.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 8304.(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(n

2、N*),记Tn=1S1+1S2+1Sn(nN*),则T2 018=()A.4 0342 018B.2 0172 018C.4 0362 019D.2 0182 0195.已知数列an满足a1=1,a2=3,an+1an-1=an(n2),则数列an的前40项和S40等于()A.20B.40C.60D.806.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 1057.(2018广东惠州模拟)数列an的前n项和为Sn,若Sn=2an-2,则数列nan的前5项和为.8.已知数列an中,a1=1,Sn为数列

3、an的前n项和,且当n2时,有2ananSn-Sn2=1成立,则S2 017=.9.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.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=.11.(2018福建福州模拟)已知数列an的前n项和为Sn,且Sn=2an-1.(1)证明:数列an是等比数列;(2)设bn=(2n-1)an,求数列bn的前n项和Tn.12.(2018河北唐山五校联考)已知数列an满足:1a1+2a2+nan=38(32n-1),nN*.(1)求数列an的通项公

4、式;(2)设bn=log3ann,求1b1b2+1b2b3+1bnbn+1.13.(2018吉林长春质量检测)等差数列an的前n项和为Sn,数列bn是等比数列,满足a1=3,b1=1,b2+S2=10,a5-2b2=a3.(1)求数列an和bn的通项公式;(2)若cn=2Sn,n为奇数,bn,n为偶数,设数列cn的前n项和为Tn,求T2n.14.(2018天津,18,13分)设an是等差数列,其前n项和为Sn(nN*);bn是等比数列,公比大于0,其前n项和为Tn(nN*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+Tn

5、)=an+4bn,求正整数n的值.答案精解精析1.Da1=S1=6,当n2时,an=Sn-Sn-1=2n+2,an=6,n=1,2n+2,n2.2.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.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,所以前60项和为S60=30

6、+230+30(30-1)24=1 830.4.C由an+2-2an+1+an=0(nN*),可得an+2+an=2an+1,所以数列an为等差数列,公差d=a2-a1=2-1=1,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.5.C由an+1=anan-1(n2),a1=1,a2=3,可得a3=3,a4=1,a5=1

7、3,a6=13,a7=1,a8=3,这是一个周期为6的数列,一个周期内的6项之和为263,又40=66+4,所以S40=6263+1+3+3+1=60.6.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.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|=765.7.答案258解析解法一:当n=1时,a1=S1=2a1-2,得a1=2,当n2时,a

8、n=Sn-Sn-1=2an-2-(2an-1-2),得an=2an-1,则数列an为等比数列,公比为2,an=2n,a1=2也符合,所以an=2n(nN*),所以nan=n2n,由错位相减法求和得前5项和T5=258.解法二:当n=1时,a1=S1=2a1-2,得a1=2,当n2时,an=Sn-Sn-1=2an-2-(2an-1-2),得an=2an-1,则数列an为等比数列,公比为2,an=2n,a1=2也符合,所以an=2n(nN*),得nan=n2n,故前5项和T5=2+222+323+424+525=2+8+24+64+160=258.8.答案11 009解析当n2时,由2ananSn

9、-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=11 009.9.答案245解析2nan=(n-1)an-1+(n+1)an+1(n2),数列nan是以a1=1为首项,2a2-a1=5为公差的等差数列,20a20=1+519=96,a20=245.10.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+

10、2.11.解析(1)证明:当n=1时,a1=S1=2a1-1,所以a1=1,当n2时,an=Sn-Sn-1=(2an-1)-(2an-1-1),所以an=2an-1,所以数列an是以1为首项,2为公比的等比数列.(2)由(1)知,an=2n-1,所以bn=(2n-1)2n-1,所以Tn=1+32+522+(2n-3)2n-2+(2n-1)2n-1,2Tn=12+322+(2n-3)2n-1+(2n-1)2n,由-得-Tn=1+2(21+22+2n-1)-(2n-1)2n=1+22-2n-121-2-(2n-1)2n=(3-2n)2n-3,所以Tn=(2n-3)2n+3.12.解析(1)当n=1

11、时,1a1=38(32-1)=3,当n2时,nan=1a1+2a2+nan-1a1+2a2+n-1an-1=38(32n-1)-38(32n-2-1)=32n-1,当n=1时,nan=32n-1也成立,所以an=n32n-1(nN*).(2)bn=log3ann=-(2n-1),因为1bnbn+1=1(2n-1)(2n+1)=1212n-1-12n+1,所以1b1b2+1b2b3+1bnbn+1=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.13.解析(1)设等差数列an的公差为d,等比数列bn的公比为q.a1=3,b1=1,b2+S2=10,a5-2b2=

12、a3,q+3+3+d=10,3+4d-2q=3+2d,d=2,q=2.an=2n+1,bn=2n-1(nN*).(2)由(1)知,Sn=n(3+2n+1)2=n(n+2).cn=1n-1n+2,n为奇数,2n-1,n为偶数.T2n=1-13+13-15+12n-1-12n+1+(21+23+25+22n-1)=2n2n+1+23(4n-1).14.解析(1)设等比数列bn的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q0,所以q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1.设等差数列an的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n,所以Sn=n(n+1)2.(2)由(1),有T1+T2+Tn=(21+22+2n)-n=2(1-2n)1-2-n=2n+1-n-2.由Sn+(T1+T2+Tn)=an+4bn可得n(n+1)2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍),或n=4.所以,n的值为4.