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1、 題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題1.設(shè)數(shù)列an的前n項(xiàng)和為Sn,滿足(1-q)Sn+qan=1,且q(q-1)0.(1)求an的通項(xiàng)公式;(2)若S3,S9,S6成等差數(shù)列,求證:a2,a8,a5成等差數(shù)列.2.已知等差數(shù)列an的首項(xiàng)a1=1,公差d=1,前n項(xiàng)和為Sn,bn=1Sn.(1)求數(shù)列bn的通項(xiàng)公式;(2)設(shè)數(shù)列bn前n項(xiàng)和為Tn,求Tn.3.已知數(shù)列an的前n項(xiàng)和Sn滿足:Sn=aa-1(an-1),a為常數(shù),且a0,a1.(1)求數(shù)列an的通項(xiàng)公式;(2)若a=13,設(shè)bn=an1+an-an+11-an+1,且數(shù)列bn的前n項(xiàng)和為Tn,求證:Tn<13.4.

2、已知等差數(shù)列an的前n項(xiàng)和為Sn,公比為q的等比數(shù)列bn的首項(xiàng)是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求數(shù)列an,bn的通項(xiàng)公式an,bn;(2)求數(shù)列1anan+1+1bnbn+1的前n項(xiàng)和Tn.5.已知數(shù)列an滿足a1=12,且an+1=an-an2(nN*).(1)證明:1anan+12(nN*);(2)設(shè)數(shù)列an2的前n項(xiàng)和為Sn,證明:12(n+2)Snn12(n+1)(nN*).6.已知數(shù)列an的首項(xiàng)為1,Sn為數(shù)列an的前n項(xiàng)和,Sn+1=qSn+1,其中q>0,nN*.(1)若2a2,a3,a2+2成等差數(shù)列,求數(shù)列an的通項(xiàng)公式;(2)設(shè)雙曲線x2

3、-y2an2=1的離心率為en,且e2=53,證明:e1+e2+en>4n-3n3n-1.參考答案題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題1.(1)解當(dāng)n=1時(shí),由(1-q)S1+qa1=1,a1=1.當(dāng)n2時(shí),由(1-q)Sn+qan=1,得(1-q)Sn-1+qan-1=1,兩式相減,得an=qan-1.又q(q-1)0,所以an是以1為首項(xiàng),q為公比的等比數(shù)列,故an=qn-1.(2)證明由(1)可知Sn=1-anq1-q,又S3+S6=2S9,所以1-a3q1-q+1-a6q1-q=2(1-a9q)1-q,化簡(jiǎn),得a3+a6=2a9,兩邊同除以q,得a2+a5=2a8.故a2,

4、a8,a5成等差數(shù)列.2.解(1)在等差數(shù)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=211×2+12×3+13×4+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?yàn)閍1=S1=aa-1(a1-1),所以a1=a.當(dāng)n2時(shí),an=Sn-Sn-1=aa-1an-aa-1an-1,得anan-1=a,所以數(shù)列an是首項(xiàng)為a,公比也為a的等比數(shù)列.所以

5、an=a·an-1=an.(2)證明當(dāng)a=13時(shí),an=13n,所以bn=an1+an-an+11-an+1=13n1+13n-13n+11-13n+1=13n+1-13n+1-1.因?yàn)?3n+1<13n,13n+1-1>13n+1,所以bn=13n+1-13n+1-1<13n-13n+1.所以Tn=b1+b2+bn<13-132+132-133+13n-13n+1=13-13n+1.因?yàn)?13n+1<0,所以13-13n+1<13,即Tn<13.4.解(1)設(shè)an公差為d,由題意得a1+2d=8,a1+2q=3,a1+d+2q=6,解得a1

6、=2,d=3,q=12,故an=3n-1,bn=12n.(2)1anan+1+1bnbn+1=131an-1an+1+1bnbn+1=131an-1an+1+22n+1,Tn=1312-15+15-18+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+13(22n+3-8)=1322n+3-13n+2-52.5.證明(1)由題意得an+1-an=-an20,即an+1an,故an12.由an=(1-an-1)an-1,得an=(1-an-1)(1-an-2)(1-a1)a1>0.由0<an12,得anan+1=anan-an2=11-an1,2,即1anan+1

7、2.(2)由題意得an2=an-an+1,所以Sn=a1-an+1.由1an+1-1an=anan+1和1anan+12,得11an+1-1an2,所以n1an+1-1a12n,因此12(n+1)an+11n+2(nN*).由得12(n+2)Snn12(n+1)(nN*).6.解(1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,兩式相減得到an+2=qan+1,n1.又由S2=qS1+1得到a2=qa1,故an+1=qan對(duì)所有n1都成立.所以,數(shù)列an是首項(xiàng)為1,公比為q的等比數(shù)列.從而an=qn-1.由2a2,a3,a2+2成等差數(shù)列,可得2a3=3a2+2,即2q2=3q+2,則(2q+1)(q-2)=0,由已知,q>0,故q=2.所以an=2n-1(nN*).(2)由(1)可知,an=qn-1.所以雙曲線x2-y2an2=1的離心率en=1+an2=1+q2(n-1).由e2=1+q2=53,解得q=43.因?yàn)?+q2(k-1)>q2(k-1),所以1+q2(k-1)>qk-1(kN*).于是e1+e2+en>1+q+qn-1=qn-1q-1,故e1+e2+en>4n-3n3n-1.