（课标专用）天津市2020高考数学二轮复习 题型练4 大题专项（二）数列的通项、求和问题

题型练4大题专项(二)数列的通项、求和问题题型练第56页1.已知等差数列an的前n项和为Sn,且a2=3,S4=16,数列bn满足a1b1+a2b2+anbn=n.(1)求bn的通项公式;(2)求数列bn+1an的前n项和Tn.解:(1)设首项为a1,公差为d的等差数列an的前n项和为Sn,且a2=3,S4=16,所以a1+d=3,4a1+432d=16,解得a1=1,d=2,所以an=1+2(n-1)=2n-1.因为a1b1+a2b2+anbn=n,所以1b1+3b2+(2n-1)bn=n,所以当n2时,1b1+3b2+(2n-3)bn-1=n-1,-得,(2n-1)bn=1,所以bn=12n-1,当n=1时,b1=1(首项符合通项),故bn=12n-1.(2)因为bn=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.已知数列an满足a1=2,an+1=2an2,nN*.(1)证明:数列1+log2an为等比数列;(2)设bn=n1+log2an,求数列bn的前n项和Sn.(1)证明由an+1=2an2,两边取以2为底的对数,得log2an+1=1+2log2an,则log2an+1+1=2(log2an+1),所以1+log2an为等比数列,首项为2,公比为2,且log2an+1=(log2a1+1)2n-1=2n.(2)解由(1)得bn=n2n.因为Sn为数列bn的前n项和,所以Sn=12+222+n2n,则12Sn=122+223+n2n+1.两式相减得12Sn=12+122+12n-n2n+1=1-12n-n2n+1,所以Sn=2-n+22n.3.已知等比数列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列bn满足b1=1,数列(bn+1-bn)an的前n项和为2n2+n.(1)求q的值;(2)求数列bn的通项公式.解:(1)由a4+2是a3,a5的等差中项,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,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列cn前n项和为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.设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.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.解:(1)设an公差为d,由题意得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.已知数列an的各项均不为零.设数列an的前n项和为Sn,数列an2的前n项和为Tn,且3Sn2-4Sn+Tn=0,nN*.(1)求a1,a2的值;(2)证明:数列an是等比数列;(3)若(-nan)(-nan+1)0对任意的nN*恒成立,求实数的所有值.(1)解3Sn2-4Sn+Tn=0,nN*,令n=1,得3a12-4a1+a12=0,因为a10,所以a1=1.令n=2,得3(1+a2)2-4(1+a2)+(1+a22)=0,即2a22+a2=0,因为a20,所以a2=-12.(2)证明因为3Sn2-4Sn+Tn=0,所以3Sn+12-4Sn+1+Tn+1=0,-得,3(Sn+1+Sn)an+1-4an+1+an+12=0.因为an+10,所以3(Sn+1+Sn)-4+an+1=0,所以3(Sn+Sn-1)-4+an=0(n2),当n2时,-,得3(an+1+an)+an+1-an=0,即an+1=-12an.因为an0,所以an+1an=-12.又由(1)知,a1=1,a2=-12,所以a2a1=-12,所以数列an是以1为首项,-12为公比的等比数列.(3)解由(2)知,an=-12n-1.因为对任意的nN*,(-nan)(-nan+1)0恒成立,所以的值介于n-12n-1和n-12n之间.因为n-12n-1n-12n0,当n为奇数时,n-12nn-12n-1恒成立,从而有n2n-1恒成立.记p(n)=n22n(n4),因为p(n+1)-p(n)=(n+1)22n+1-n22n=-n2+2n+12n+10不符合题意.若0,当n为奇数时,n-12nn-12n-1恒成立,从而有-n2n恒成立.由(*)式知,当n5,且n-1时,有-1nn2n,所以0,nN*.(1)若2a2,a3,a2+2成等差数列,求数列an的通项公式;(2)设双曲线x2-y2an2=1的离心率为en,且e2=53,证明:e1+e2+en4n-3n3n-1.(1)解由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,两式相减得到an+2=qan+1,n1.又由S2=qS1+1得到a2=qa1,故an+1=qan对所有n1都成立.所以,数列an是首项为1,公比为q的等比数列.从而an=qn-1.由2a2,a3,a2+2成等差数列,可得2a3=3a2+2,即2q2=3q+2,则(2q+1)(q-2)=0,由已知,q0,故q=2.所以an=2n-1(nN*).(2)证明由(1)可知,an=qn-1.所以双曲线x2-y2an2=1的离心率en=1+an2=1+q2(n-1).由e2=1+q2=53,解得q=43.因为1+q2(k-1)q2(k-1),所以1+q2(k-1)qk-1(kN*).于是e1+e2+en1+q+qn-1=qn-1q-1,故e1+e2+en4n-3n3n-1.6