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第一讲等差数列、等比数列(40分钟70分)一、选择题(每小题5分,共25分)1.设Sn,Tn分别是等差数列an,bn的前n项和,若SnTn=n2n+1 (nN*),则a5b5=()A.513B.923C.1123D.919【解析】选D.a5b5=S9T9=919.2.等差数列an的公差为2,若a2,a4,a8成等比数列,则an的前n项和Sn=()A.n(n+1)B.n(n-1)C.n(n+1)2D.n(n-1)2【解析】选A.由a42=a2a8得(a1+3d)2=(a1+d)(a1+7d),所以(a1+6)2=(a1+2)(a1+14),解得a1=2所以Sn=na1+n(n-1)2d=n2+n.3.已知数列an满足:an+1an+1+1=12,且a2=2,则a4等于()A.-12B.23C.12D.11【解析】选D.因为数列an满足:an+1an+1+1=12,所以an+1+1=2(an+1),即数列an+1是等比数列,公比为2.则a4+1=22(a2+1)=12,解得a4=11.4.已知数列an是公差不为0的等差数列,a2=3,且a3,a5,a8成等比数列,设bn=2anan+1,则数列bn的前n项和Tn为()A.n-1n B.nn+2 C.2n2n+1 D.n2n+4【解析】选B.设公差为d(d0),首项为a1,所以a1+d=3,(a1+2d)(a1+7d)=(a1+4d)2,解得a1=2,d=1,所以an=n+1,bn=2(n+1)(n+2)=21n+1-1n+2,所以Tn=212-13+213-14+21n+1-1n+2=212-1n+2=nn+2.5.记Sn为等差数列an的前n项和,若3S3=S2+S4,a1=2,则a5=()A.-12B. -10C.10D.12【解析】选B.33a1+322d=2a1+d+4a1+432d9a1+9d=6a1+7d3a1+2d=06+2d=0d=-3,所以a5=a1+4d=2+4(-3)=-10.二、填空题(每小题5分,共15分)6.若an为等比数列,an0,且a2 018=22,则1a2 017+2a2 019的最小值为_.【解析】1a2 017+2a2 019=a2 019+2a2017a2 018222a2 018a2 0182=4答案:47.设Sn为等差数列an的前n项和,满足S2=S6,S55-S44=2,则a1=_,公差d=_.【解析】因为Sn为等差数列an的前n项和,满足S2=S6,S55-S44=2,所以2a1+d=6a1+15d,5a1+10d5-4a1+6d4=2,解得a1=-14,d=4.答案:-1448.对给定的正整数n(n6),定义f(x)=a0+a1x+a2x2+anxn,其中a0=1,ai=2ai-1(iN*,in),则a6=_;当n=2 017时,f(2)=_.【解析】因为a0=1,ai=2ai-1(iN*,in),所以a6=2a5=22a4=26a0=64.f(2)=20+212+2222+2323+22 01722 017=1-42 0181-4=42 018-13.答案:6442 018-13三、解答题(每小题10分,共30分)9.等比数列an的各项均为正数,且2a1+3a2=1,a32=9a2a6,(1)求数列an的通项公式.(2)设bn=log3a1+log3a2+log3an,求数列1bn的前n项和.【解析】(1)设等比数列an的公比为q,由a32=9a2a6,得a32=9a42,所以q2=19,由条件可知q0,故q=13.由2a1+3a2=1得2a1+3a1q=1,所以a1=13.故数列an的通项公式为an=13n.(2)bn=log3a1+log3a2+log3an=-(1+2+n)=-n(n+1)2,故1bn=-2n(n+1)=-21n-1n+1.1b1+1b2+1bn=-21-12+12-13+1n-1n+1=-2nn+1.10.设数列an的前n项和为Sn,数列Sn的前n项和为Tn,满足Tn=2Sn-n2,nN*.(1)求a1,a2,a3的值.(2)求数列an的通项公式.【解析】(1)因为S1=T1=2S1-1,S1=1=a1,所以a1=1.因为S1+S2=T2=2S2-4,所以a2=4.因为S1+S2+S3=T3=2S3-9,所以a3=10.(2)因为Tn=2Sn-n2 ,Tn-1=2Sn-1-(n-1)2,所以-得,Sn=2an-2n+1(n2),因为S1=2a1-21+1,所以Sn=2an-2n+1(n1) , Sn-1=2an-1-2n+3,-得,an=2an-1+2(n2)an+2=2(an-1+2).因为a1+2=3,所以an+2是首项为3,公比为2的等比数列,an+2=32n-1,故an=32n-1-2.11.设数列an的前n项和为Sn,且Sn=n2-n+1,正项等比数列bn的前n项和为Tn,且b2=a2,b4=a5.(1)求an和bn的通项公式.(2)数列cn中,c1=a1,且cn=cn+1-Tn,求cn的通项公式cn.【解析】(1)因为Sn=n2-n+1,所以令n=1,a1=1,an=Sn-Sn-1=2(n-1),(n2),经检验a1=1不能与an(n2)合并,所以an=1,n=1,2(n-1),n2.又因为数列bn为正项等比数列,b2=a2=2,b4=a5=8,所以b4b2=q2=4,所以q=2,所以b1=1,所以bn=2n-1.(2)Tn=1-2n1-2=2n-1,因为c2-c1=21-1,c3-c2=22-1,cn-cn-1=2n-1-1,以上各式相加得cn-c1=2(1-2n-1)1-2-(n-1),c1=a1=1,所以cn-1=2n-n-1,所以cn=2n-n.(20分钟20分)1.(10分)已知数列an中,a2=1,前n项和为Sn,且Sn=n(an-a1)2.(1)求a1,a3.(2)求证:数列an为等差数列,并写出其通项公式.【解析】(1)令n=1,则a1=S1=1(a1-a1)2=0,令n=3,则S3=3(a3-a1)2,即0+1+a3=3a32,解得a3=2.(2)由Sn=n(an-a1)2,即Sn=nan2,得Sn+1=(n+1)an+12,-,得(n-1)an+1=nan,于是,nan+2=(n+1)an+1,-,得nan+2+nan=2nan+1,即an+2+an=2an+1,又a1=0,a2=1,a2-a1=1,所以数列an是以0为首项,1为公差的等差数列.所以an=n-1.【提分备选】已知正项数列an的前n项和为Sn,且a1=1,Sn+1+Sn=an+12,数列bn满足bnbn+1=3an,且b1=1.(1)求数列an,bn的通项公式.(2)记Tn=anb2+an-1b4+a1b2n,求Tn.【解析】(1)因为Sn+1+Sn=an+12,Sn+Sn-1=an2(n2),-得:an+1+an=an+12-an2,所以(an+1+an)(an+1-an-1)=0,因为an+10,an0,所以an+1+an0,所以an+1-an=1(n2),又由S2+S1=a22得2a1+a2=a22,即a22-a2-2=0,所以a2=2,a2=-1(舍去),所以a2-a1=1,所以an是以1为首项,1为公差的等差数列,所以an=n.又因为bnbn+1=3an=3n,所以bn-1bn=3n-1(n2),得:bn+1bn-1=3(n2),又由b1=1,可求b2=3,故b1,b3,b2n-1是首项为1,公比为3的等比数列,b2,b4,b2n是首项为3,公比为3的等比数列.所以b2n-1=3n-1,b2n=33n-1=3n.所以bn=3n-12(n为奇数),3n2(n为偶数).(2)由(1)得:Tn=3an+32an-1+33an-2+3na1,3Tn=32an+33an-1+34an-2+3n+1a1, -得:2Tn=-3an+32(an-an-1)+33(an-1-an-2)+3n(a2-a1)+3n+1a1,由an=n,所以2Tn=-3n+32+33+3n+3n+1=-3n+32(1-3n)1-3=-3n-92+123n+2,所以Tn=3n+24-3n2-94.2.(10分)(2018日照一模)已知数列an前n项和Sn满足:2Sn+an=1.(1)求数列an的通项公式.(2)设bn=2an+1(1+an)(1+an+1),数列bn的前n项和为Tn,求证:Tn14.【解析】(1)因为2Sn+an=1,所以2Sn+1+an+1=1,两式相减可得2an+1+an+1-an=0,即3an+1=an,即an+1an=13,又2S1+a1=1,所以a1=13,所以数列an是公比为13的等比数列.故an=1313n-1=13n,数列an的通项公式为an=13n. (2)因为bn=2an+1(1+an)(1+an+1),所以bn=213n+11+13n1+13n+1=23n+13n+13n3n+1+13n+1=23n(3n+1)(3n+1+1)=13n+1-13n+1+1所以Tn=b1+b2+bn=131+1-132+1+132+1-133+1+13n+1-13n+1+1=14-13n+1+114.所以Tn14.
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