（課標(biāo)專用）天津市2020高考數(shù)學(xué)二輪復(fù)習(xí) 題型練4 大題專項(xiàng)（二）數(shù)列的通項(xiàng)、求和問題

題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問題題型練第56頁 1.已知等差數(shù)列an的前n項(xiàng)和為Sn,且a2=3,S4=16,數(shù)列bn滿足a1b1+a2b2+anbn=n.(1)求bn的通項(xiàng)公式;(2)求數(shù)列bn+1an的前n項(xiàng)和Tn.解:(1)設(shè)首項(xiàng)為a1,公差為d的等差數(shù)列an的前n項(xiàng)和為Sn,且a2=3,S4=16,所以a1+d=3,4a1+4×32d=16,解得a1=1,d=2,所以an=1+2(n-1)=2n-1.因?yàn)閍1b1+a2b2+anbn=n,所以1·b1+3·b2+(2n-1)bn=n,所以當(dāng)n2時(shí),1·b1+3·b2+(2n-3)bn-1=n-1,-得,(2n-1)bn=1,所以bn=12n-1,當(dāng)n=1時(shí),b1=1(首項(xiàng)符合通項(xiàng)),故bn=12n-1.(2)因?yàn)閎n=12n-1,所以bn+1an=12n+12n-1=1(2n-1)(2n+1)=1212n-1-12n+1,所以Tn=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.2.已知數(shù)列an滿足a1=2,an+1=2an2,nN*.(1)證明:數(shù)列1+log2an為等比數(shù)列;(2)設(shè)bn=n1+log2an,求數(shù)列bn的前n項(xiàng)和Sn.(1)證明由an+1=2an2,兩邊取以2為底的對(duì)數(shù),得log2an+1=1+2log2an,則log2an+1+1=2(log2an+1),所以1+log2an為等比數(shù)列,首項(xiàng)為2,公比為2,且log2an+1=(log2a1+1)×2n-1=2n.(2)解由(1)得bn=n2n.因?yàn)镾n為數(shù)列bn的前n項(xiàng)和,所以Sn=12+222+n2n,則12Sn=122+223+n2n+1.兩式相減得12Sn=12+122+12n-n2n+1=1-12n-n2n+1,所以Sn=2-n+22n.3.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.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+7·12+3.設(shè)Tn=3+7·12+11·122+(4n-5)·12n-2,n2,12Tn=3·12+7·122+(4n-9)·12n-2+(4n-5)·12n-1,所以12Tn=3+4·12+4·122+4·12n-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.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.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=13(12-15)+15-18+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+13(22n+3-8)=1322n+3-13n+2-52.5.已知數(shù)列an的各項(xiàng)均不為零.設(shè)數(shù)列an的前n項(xiàng)和為Sn,數(shù)列an2的前n項(xiàng)和為Tn,且3Sn2-4Sn+Tn=0,nN*.(1)求a1,a2的值;(2)證明:數(shù)列an是等比數(shù)列;(3)若(-nan)(-nan+1)<0對(duì)任意的nN*恒成立,求實(shí)數(shù)的所有值.(1)解3Sn2-4Sn+Tn=0,nN*,令n=1,得3a12-4a1+a12=0,因?yàn)閍10,所以a1=1.令n=2,得3(1+a2)2-4(1+a2)+(1+a22)=0,即2a22+a2=0,因?yàn)閍20,所以a2=-12.(2)證明因?yàn)?Sn2-4Sn+Tn=0,所以3Sn+12-4Sn+1+Tn+1=0,-得,3(Sn+1+Sn)an+1-4an+1+an+12=0.因?yàn)閍n+10,所以3(Sn+1+Sn)-4+an+1=0,所以3(Sn+Sn-1)-4+an=0(n2),當(dāng)n2時(shí),-,得3(an+1+an)+an+1-an=0,即an+1=-12an.因?yàn)閍n0,所以an+1an=-12.又由(1)知,a1=1,a2=-12,所以a2a1=-12,所以數(shù)列an是以1為首項(xiàng),-12為公比的等比數(shù)列.(3)解由(2)知,an=-12n-1.因?yàn)閷?duì)任意的nN*,(-nan)(-nan+1)<0恒成立,所以的值介于n-12n-1和n-12n之間.因?yàn)閚-12n-1·n-12n<0對(duì)任意的nN*恒成立,所以=0適合.若>0,當(dāng)n為奇數(shù)時(shí),n-12n<<n-12n-1恒成立,從而有<n2n-1恒成立.記p(n)=n22n(n4),因?yàn)閜(n+1)-p(n)=(n+1)22n+1-n22n=-n2+2n+12n+1<0,所以p(n)p(4)=1,即n22n1,所以n2n1n,(*)從而當(dāng)n5,且n2時(shí),有2nn2n-1,所以>0不符合題意.若<0,當(dāng)n為奇數(shù)時(shí),n-12n<<n-12n-1恒成立,從而有-<n2n恒成立.由(*)式知,當(dāng)n5,且n-1時(shí),有-1nn2n,所以<0不符合題意.綜上,實(shí)數(shù)的所有值為0.6.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-y2an2=1的離心率為en,且e2=53,證明:e1+e2+en>4n-3n3n-1.(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-1.6