高考數(shù)學(xué)復(fù)習(xí)：第五章 ：第四節(jié)

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1、 精品資料課時(shí)提升作業(yè)(三十三)一、選擇題1.(2013南昌模擬)已知等比數(shù)列an公比為q,其前n項(xiàng)和為Sn,若S3,S9,S6成等差數(shù)列,則q3等于()(A)-12(B)1(C)-12或1(D)-1或122.(2013長春模擬)在等差數(shù)列an中,a9=12a12+6,則數(shù)列an的前11項(xiàng)和S11等于()(A)24(B)48(C)66(D)1323.已知數(shù)列an的通項(xiàng)公式是an=2n-3(15)n,則其前20項(xiàng)和為()(A)380-35(1-1519)(B)400-25(1-1520)(C)420-34(1-1520)(D)440-45(1-1520)4.(2013阜陽模擬)已知直線(3m+1

2、)x+(1-m)y-4=0所過定點(diǎn)的橫、縱坐標(biāo)分別是等差數(shù)列an的第一項(xiàng)與第二項(xiàng),若bn=1anan+1,數(shù)列bn的前n項(xiàng)和為Tn,則T10=()(A)921(B)1021(C)1121(D)20215.(2013太原模擬)已知等差數(shù)列an的公差d0,且a1,a3,a9成等比數(shù)列,則a1+a3+a9a2+a4+a10=()(A)914(B)1115(C)1316(D)15176.數(shù)列an的前n項(xiàng)和Sn=3n+b(b是常數(shù)),若這個(gè)數(shù)列是等比數(shù)列,那么b為() (A)3(B)0(C)-1(D)17.等差數(shù)列an的前n項(xiàng)和為Sn,已知am-1+am+1-am2=0,S2m-1=38,則m=()(A

3、)38(B)20(C)10(D)98.(能力挑戰(zhàn)題)數(shù)列an的前n項(xiàng)和Sn=2n-1,則a12+a22+a32+an2等于()(A)(2n-1)2(B)13(2n-1)2(C)4n-1(D)13(4n-1)二、填空題9.已知等差數(shù)列an的前n項(xiàng)和為Sn.若a3=20-a6,則S8等于.10.數(shù)列1+2n-1的前n項(xiàng)和為.11.(2013蕪湖模擬)已知數(shù)列an中,a1=1,a2=2,當(dāng)整數(shù)n1時(shí),Sn+1+Sn-1=2(Sn+S1)都成立,則S5=.12.(2013哈爾濱模擬)在數(shù)列an中,若對任意的n均有an+an+1+an+2為定值(nN+),且a7=2,a9=3,a98=4,則此數(shù)列an的

4、前100項(xiàng)的和S100=.三、解答題13.已知數(shù)列l(wèi)og2(an-1)(nN+)為等差數(shù)列,且a1=3,a3=9.(1)求數(shù)列an的通項(xiàng)公式.(2)求和:Sn=1a2-a1+1a3-a2+1an+1-an.14.(2012湖州模擬)設(shè)an是等差數(shù)列,bn是各項(xiàng)都為正數(shù)的等比數(shù)列,且a1=b1=1,a3+b5=21,a5+b3=13.(1)求an,bn的通項(xiàng)公式.(2)求數(shù)列anbn的前n項(xiàng)和Sn.15.(能力挑戰(zhàn)題)已知數(shù)列an的通項(xiàng)公式是an=n2n-1,bn=an+2anan+1,求數(shù)列bn的前n項(xiàng)和. 答案解析1.【解析】選A.當(dāng)q=1時(shí),顯然不可能;當(dāng)q1時(shí),根據(jù)已知得2a1(1-q9

5、)1-q=a1(1-q3)1-q+a1(1-q6)1-q,即2q9=q6+q3,即2q6-q3-1=0,解得q3=1(舍),或q3=-12.2.【解析】選D.設(shè)公差為d,則a1+8d=12a1+112d+6,即a1+5d=12,即a6=12,所以S11=11a6=132.3.【解析】選C.由an=2n-3(15)n,得S20=2(1+2+20)-3(15+152+1520)=220(1+20)2-315(1-1520)1-15=420-34(1-1520),故選C.4.【解析】選B.將直線方程化為(x+y-4)+m(3x-y)=0,令x+y-4=0,3x-y=0,解得x=1,y=3,即直線過定

6、點(diǎn)(1,3),所以a1=1,a2=3,公差d=2,an=2n-1,bn=1anan+1=12(12n-1-12n+1),T10=12(11-13+13-15+120-1-120+1)=12(11-121)=1021.5.【解析】選C.等差數(shù)列an中,a1=a1,a3=a1+2d,a9=a1+8d,因?yàn)閍1,a3,a9恰好構(gòu)成某等比數(shù)列,所以有a32=a1a9,即(a1+2d)2=a1(a1+8d),解得d=a1,所以該等差數(shù)列的通項(xiàng)為an=nd.則a1+a3+a9a2+a4+a10的值為1316.6.【思路點(diǎn)撥】根據(jù)數(shù)列的前n項(xiàng)和減去前n-1項(xiàng)的和得到數(shù)列的第n項(xiàng)的通項(xiàng)公式,即可得到此等比數(shù)列

7、的首項(xiàng)與公比,根據(jù)首項(xiàng)和公比,利用等比數(shù)列的前n項(xiàng)和公式表示出前n項(xiàng)的和,與已知的Sn=3n+b對比后,即可得到b的值.【解析】選C.因?yàn)閍n=Sn-Sn-1=(3n+b)-(3n-1+b)=3n-3n-1=23n-1(n2),所以此數(shù)列是首項(xiàng)為2,公比為3的等比數(shù)列,則Sn=2(1-3n)1-3=3n-1,所以b=-1. 7.【解析】選C.因?yàn)閍n是等差數(shù)列,所以am-1+am+1=2am,由am-1+am+1-am2=0,得2am-am2=0,所以am=2(am=0舍),又S2m-1=38,即(2m-1)(a1+a2m-1)2=38,即(2m-1)2=38,解得m=10,故選C.8.【解析

8、】選D.an=Sn-Sn-1=2n-1(n1),又a1=S1=1=20,適合上式,an=2n-1(nN+),an2是a12=1,q=22的等比數(shù)列,由求和公式得a12+a22+a32+an2=1(1-4n)1-4=13(4n-1).9.【解析】因?yàn)閍3=20-a6,所以S8=4(a3+a6)=420=80.答案：8010.【解析】前n項(xiàng)和Sn=1+20+1+21+1+22+1+2n-1=n+1-2n1-2=n+2n-1.答案：n+2n-111.【解析】由Sn+1+Sn-1=2(Sn+S1)得(Sn+1-Sn)-(Sn-Sn-1)=2S1=2,即an+1-an=2(n2),數(shù)列an從第二項(xiàng)起構(gòu)成

9、等差數(shù)列,則S5=1+2+4+6+8=21.答案：2112.【解析】設(shè)定值為M,則an+an+1+an+2=M,進(jìn)而an+1+an+2+an+3=M,后式減去前式得an+3=an,即數(shù)列an是以3為周期的數(shù)列.由a7=2,可知a1=a4=a7=a100=2,共34項(xiàng),其和為68;由a9=3,可得a3=a6=a99=3,共33項(xiàng),其和為99;由a98=4,可得a2=a5=a98=4,共33項(xiàng),其和為132.故數(shù)列an的前100項(xiàng)的和S100=68+99+132=299.答案：29913.【解析】(1)設(shè)等差數(shù)列l(wèi)og2(an-1)的公差為d.由a1=3,a3=9得2(log22+d)=log22

10、+log28,即d=1.所以log2(an-1)=1+(n-1)1=n,即an=2n+1.(2)因?yàn)?an+1-an=12n+1-2n=12n,所以Sn=1a2-a1+1a3-a2+1an+1-an=121+122+123+12n=12-12n121-12=1-12n.14.【解析】(1)設(shè)an的公差為d,bn的公比為q,則依題意有q0且1+2d+q4=21,1+4d+q2=13,解得d=2,q=2.所以an=1+(n-1)d=2n-1,bn=qn-1=2n-1.(2)anbn=2n-12n-1,Sn=1+321+522+2n-32n-2+2n-12n-1,2Sn=2+3+52+2n-32n-

11、3+2n-12n-2.-,得Sn=2+2+22+222+22n-2-2n-12n-1=2+2(1+12+122+12n-2)-2n-12n-1=2+21-12n-11-12-2n-12n-1=6-2n+32n-1.【變式備選】已知各項(xiàng)都不相等的等差數(shù)列an的前6項(xiàng)和為60,且a6為a1和a21的等比中項(xiàng).(1)求數(shù)列an的通項(xiàng)公式.(2)若數(shù)列bn滿足bn+1-bn=an(nN+),且b1=3,求數(shù)列1bn的前n項(xiàng)和Tn.【解析】(1)設(shè)等差數(shù)列an的公差為d(d0),則6a1+15d=60,a1(a1+20d)=(a1+5d)2,解得d=2,a1=5,an=2n+3.(2)由bn+1-bn=

12、an,bn-bn-1=an-1(n2,nN+),bn=(bn-bn-1)+(bn-1-bn-2)+(b2-b1)+b1=an-1+an-2+a1+b1=n(n+2),當(dāng)n=1時(shí),b1=3也適合上式, bn=n(n+2)(nN+).1bn=1n(n+2)=12(1n-1n+2),Tn=12(1-13+12-14+1n-1n+2)=12(32-1n+1-1n+2)=3n2+5n4(n+1)(n+2).15.【解析】ak+2akak+1=(k+2)2k+1k2k-1(k+1)2k=k+2k(k+1)2k-2=2(k+1)-kk(k+1)2k-2=1k2k-3-1(k+1)2k-2,k=1,2,3,n

13、 故a3a1a2+a4a2a3+an+2anan+1=(112-2-122-1)+(122-1-1320)+1n2n-3-1(n+1)2n-2=112-2-1(n+1)2n-2=4-1(n+1)2n-2.【方法技巧】裂項(xiàng)相消法的應(yīng)用技巧裂項(xiàng)相消法的基本思想是把數(shù)列的通項(xiàng)an分拆成an=bn+1-bn或者an=bn-bn+1或者an=bn+2-bn等,從而達(dá)到在求和時(shí)逐項(xiàng)相消的目的,在解題中要善于根據(jù)這個(gè)基本思想變換數(shù)列an的通項(xiàng)公式,使之符合裂項(xiàng)相消的條件.在裂項(xiàng)時(shí)一定要注意把數(shù)列的通項(xiàng)分拆成的兩項(xiàng)一定是某個(gè)數(shù)列中的相鄰的兩項(xiàng)或者是等距離間隔的兩項(xiàng),只有這樣才能實(shí)現(xiàn)逐項(xiàng)相消后剩下幾項(xiàng),達(dá)到求和的目的.