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高考大题专项三高考中的数列1.(2018山西吕梁一模,17)已知an是首项为1的等比数列,数列bn满足b1=2,b2=5,且anbn+1=anbn+an+1.(1)求数列an的通项公式;(2)求数列bn的前n项和.2.(2018福建龙岩4月质检,17)已知正项等比数列an的前n项和为Sn,且Sn=2an-1(nN+).(1)求数列an的通项公式;(2)若bn=lgan,求数列an+bn的前n项和Tn.3.(2018北京海淀期末,15)已知等差数列an的前n项和Sn,且a2=5,S3=a7.(1)数列an的通项公式;(2)若bn=2an,求数列an+bn的前n项和.4.(2018河北唐山一模,17)已知数列an为单调递增数列,Sn为其前n项和,2Sn=an2+n.(1)求an的通项公式;(2)若bn=an+22n+1anan+1,Tn为数列bn的前n项和,证明:Tn12.5.(2018湖南衡阳二模,17)等差数列an中,a3=1,a7=9,Sn为等比数列bn的前n项和,且b1=2,若4S1,3S2,2S3成等差数列.(1)求数列an,bn的通项公式;(2)设cn=|an|bn,求数列cn的前n项和Tn.6.已知数列an的前n项和为Sn,Sn=(m+1)-man对任意的nN+都成立,其中m为常数,且m-1.(1)求证:数列an是等比数列;(2)记数列an的公比为q,设q=f(m),若数列bn满足b1=a1,bn=f(bn-1)(n2,nN+).求证:数列1bn是等差数列;(3)在(2)的条件下,设cn=bnbn+1,数列cn的前n项和为Tn,求证:Tn1.7.(2018宿州十三所中学期中,17)已知数列an的前n项和为Sn,并且满足a1=1,nan+1=Sn+n(n+1).(1)求数列an的通项公式;(2)若bn=an2n,数列bn的前n项和为Tn,求Tn;(3)在(2)的条件下,是否存在常数,使得数列Tn+an+2为等比数列?若存在,试求出;若不存在,说明理由.参考答案高考大题专项三高考中的数列1.解 (1)把n=1代入已知等式得a1b2=a1b1+a2,a2=a1b2-a1b1=3a1.an是首项为1,公比为3的等比数列,即an=3n-1.(2)由已知得bn+1-bn=an+1an=3,bn是首项为2,公差为3的等差数列,其通项公式为bn=3n-1,Sn=n(b1+bn)2=n(2+3n-1)2=3n2+n2.2.解 (1)由Sn=2an-1(nN+),可得S1=2a1-1,a1=2a1-1,a1=1.又S2=2a2-1,a1+a2=2a2-1,a2=2.数列an是等比数列,公比q=a2a1=2,数列an的通项公式为an=2n-1.(2)由(1)知,bn=lgan=(n-1)lg 2,Tn=(b1+a1)+(b2+a2)+(bn+an)=(0+1)+(lg 2+2)+(n-1)lg 2+2n-1=lg 2+2lg 2+(n-1)lg 2+(1+2+2n-1)=n(n-1)2lg 2+2n-1.3.解 (1)设等差数列an的首项为a1,公差为d,则a1+d=5,3a1+3d=a1+6d,解得a1=3,d=2.由an=a1+(n-1)d,则an=2n+1.因此,通项公式为an=2n+1.(2)由(1)可知:an=2n+1,则bn=22n+1,bn+1bn=22(n+1)+122n+1=4.因为b1=23=8,所以bn是首项为8,公比为q=4的等比数列.记an+bn的前n项和为Tn,则Tn=(a1+b1)+(a2+b2)+(an+bn)=(a1+a2+an)+(b1+b2+bn)=n(a1+an)2+b1(1-qn)1-q=n2+2n+8(4n-1)3.4.(1)解当n=1时,2S1=2a1=a12+1.所以(a1-1)2=0,即a1=1.又an为单调递增数列,所以an1.由2Sn=an2+n得2Sn+1=an+12+n+1,所以2Sn+1-2Sn=an+12-an2+1,整理得2an+1=an+12-an2+1,即an2=(an+1-1)2,所以an=an+1-1,即an+1-an=1,所以an是以1为首项,1为公差的等差数列,所以an=n.(2)证明因为bn=an+22n+1anan+1=n+22n+1n(n+1)=12nn-12n+1(n+1),所以Tn=1211-1222+1222-1233+12nn-12n+1(n+1)=1211-12n+1(n+1)0,Tn=10+123+324+(2n-7)2n-1+(2n-5)2n,2Tn=20+124+325+(2n-7)2n+(2n-5)2n+1,由-,得-Tn=-10+8+2(24+2n)-(2n-5)2n+1,Tn=34+(2n-7)2n+1.Tn=6,n=1,10,n=2,34+(2n-7)2n+1,n3.6.证明 (1)当n=1时,a1=S1=1.Sn=(m+1)-man,Sn-1=(m+1)-man-1(n2),由-,得an=man-1-man(n2),即(m+1)an=man-1.a10,m-1,an-10,m+10.anan-1=mm+1(n2).数列an是首项为1,公比为mm+1的等比数列.(2)f(m)=mm+1,b1=a1=1,bn=f(bn-1)=bn-1bn-1+1(n2),1bn=bn-1+1bn-1(n2),1bn-1bn-1=1(n2),数列1bn是首项为1,公差为1的等差数列.(3)由(2)得1bn=n,则bn=1n,故cn=bnbn+1=1n(n+1),因此,Tn=112+123+1n(n+1)=11-12+12-13+13-14+1n-1n+1=1-1n+11.7.解 (1)nan+1=Sn+n(n+1),当n2时,(n-1)an=Sn-1+n(n-1),由-可得an+1-an=2(n2),且a1=1,a2=S1+1(1+1)=3,数列an是首项为1,公差为2的等差数列,即an=2n-1.(2)由(1)知数列an=2n-1,bn=2n-12n,则Tn=121+322+523+2n-32n-1+2n-12n,12Tn=122+323+524+2n-32n+2n-12n+1,由-得,12Tn=12+2122+123+12n-2n-12n+1=12+214(1-12n-1)1-12-2n-12n+1,Tn=3-2n+32n.(3)由(2)知Tn=3-2n+32n,Tn+an+2=(3-2n+32n+)2n+3=(3+)2n+3-12n,要使数列Tn+an+2为等比数列,当且仅当3+=0,即=-3.故存在=-3,使得数列Tn+an+2为等比数列.
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