1、题型练4大题专项(二)数列的通项、求和问题1.(2015河北唐山高三一模)设数列an的前n项和为Sn,满足(1-q)Sn+qan=1,且q(q-1)0.(1)求an的通项公式;(2)若S3,S9,S6成等差数列,求证:a2,a8,a5成等差数列.2.已知等差数列an的首项a1=1,公差d=1,前n项和为Sn,bn=1Sn.(1)求数列bn的通项公式;(2)设数列bn前n项和为Tn,求Tn.3.(2015广东广州调研)已知数列an的前n项和Sn满足:Sn=aa-1(an-1),a为常数,且a0,a1.(1)求数列an的通项公式;(2)若a=13,设bn=an1+an-an+11-an+1,且数列
2、bn的前n项和为Tn,求证:Tn13.4.已知等差数列an的前n项和为Sn,公比为q的等比数列bn的首项是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列an,bn的通项公式an,bn;(2)求数列1anan+1+1bnbn+1的前n项和Tn.5.(2015浙江高考)已知数列an满足a1=12且an+1=an-an2(nN*).(1)证明:1anan+12(nN*);(2)设数列an2的前n项和为Sn,证明:12(n+2)Snn12(n+1)(nN*).6.设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn2-(n2+n-3)Sn-3(n2+n)=0,nN*.(1)求a
3、1的值;(2)求数列an的通项公式;(3)证明:对一切正整数n,有1a1(a1+1)+1a2(a2+1)+1an(an+1)13.参考答案1.解:(1)当n=1时,由(1-q)S1+qa1=1,a1=1.当n2时,由(1-q)Sn+qan=1,得(1-q)Sn-1+qan-1=1,两式相减,得an=qan-1.又q(q-1)0,所以an是以1为首项,q为公比的等比数列,故an=qn-1.(2)由(1)可知Sn=1-anq1-q,又S3+S6=2S9,所以1-a3q1-q+1-a6q1-q=2(1-a9q)1-q,化简,得a3+a6=2a9,两边同除以q,得a2+a5=2a8.故a2,a8,a5
4、成等差数列.2.解:(1)在等差数列an中,a1=1,公差d=1,Sn=na1+n(n-1)2d=n2+n2,bn=2n2+n.(2)bn=2n2+n=2n(n+1)=21n-1n+1,Tn=b1+b2+b3+bn=2112+123+134+1n(n+1)=21-12+12-13+13-14+1n-1n+1=21-1n+1=2nn+1.故Tn=2nn+1.3.(1)解:因为a1=S1=aa-1(a1-1),所以a1=a.当n2时,an=Sn-Sn-1=aa-1an-aa-1an-1,得anan-1=a,所以数列an是首项为a,公比也为a的等比数列.所以an=aan-1=an.(2)证明:当a=
5、13时,an=13n,所以bn=an1+an-an+11-an+1=13n1+13n-13n+11-13n+1=13n+1-13n+1-1.因为13n+113n+1,所以bn=13n+1-13n+1-113n-13n+1.所以Tn=b1+b2+bn13-132+132-133+13n-13n+1=13-13n+1.因为-13n+10,所以13-13n+113,即Tn0.由0an12,得anan+1=anan-an2=11-an1,2,即1anan+12.(2)由题意得an2=an-an+1,所以Sn=a1-an+1.由1an+1-1an=anan+1和1anan+12,得11an+1-1an2
6、,所以n1an+1-1a12n,因此12(n+1)an+11n+2(nN*).由得12(n+2)Snn12(n+1)(nN*).6.(1)解:由题意知Sn2-(n2+n-3)Sn-3(n2+n)=0,nN*.令n=1,有S12-(12+1-3)S1-3(12+1)=0,可得S12+S1-6=0,解得S1=-3或2,即a1=-3或2.又an为正数,所以a1=2.(2)解:由Sn2-(n2+n-3)Sn-3(n2+n)=0,nN*,得(Sn+3)(Sn-n2-n)=0,则Sn=n2+n或Sn=-3.又数列an的各项均为正数,所以Sn=n2+n,Sn-1=(n-1)2+(n-1),所以当n2时,an=Sn-Sn-1=n2+n-(n-1)2+(n-1)=2n.又a1=2=21,所以an=2n.(3)证明:当n=1时,1a1(a1+1)=123=1613成立.当n2时,1an(an+1)=12n(2n+1)1(2n-1)(2n+1)=1212n-1-12n+1,所以1a1(a1+1)+1a2(a2+1)+1an(an+1)16+1213-15+12n-1-12n+1=16+1213-12n+116+16=13.所以对一切正整数n,有1a1(a1+1)+1a2(a2+1)+1an(an+1)13.
