（新課標(biāo)）天津市2019年高考數(shù)學(xué)二輪復(fù)習(xí) 題型練4 大題專(zhuān)項(xiàng)（二）數(shù)列的通項(xiàng)、求和問(wèn)題 理.doc

題型練4大題專(zhuān)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(mǎn)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T(mén)n,求Tn.3.(2018浙江,20)已知等比數(shù)列an的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列bn滿(mǎn)足b1=1,數(shù)列(bn+1-bn)an的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式.4.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(mǎn)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(xiàn)x2-y2an2=1的離心率為en,且e2=53,證明:e1+e2+en>4n-3n3n-1.題型練4大題專(zhuān)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,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=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.解 (1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因?yàn)閝>1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn前n項(xiàng)和為Sn,由cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1.故bn-bn-1=(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.設(shè)Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12n-2+(4n-5)12n-1,所以12Tn=3+412+4122+412n-2-(4n-5)12n-1,因此Tn=14-(4n+3)12n-2,n2,又b1=1,所以bn=15-(4n+3)12n-2.4.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=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+12.(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.(1)解 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)證明 由(1)可知,an=qn-1.所以雙曲線(xiàn)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.