高优指导2021版高考数学一轮复习大题专项练3高考中的数列文北师大版.doc

高考大题专项练3高考中的数列高考大题专项练第6页1.(2015大连一模)等差数列an的前n项和为Sn,等比数列bn的公比为,满足S3=15,a1+2b1=3,a2+4b2=6.(1)求数列an,bn的通项公式an,bn;(2)求数列anbn的前n项和Tn.解:(1)设an的公差为d,所以解得a1=2,d=3,b1=,所以an=3n-1,bn=.(2)由(1)知Tn=2+5+8+(3n-4)+(3n-1),Tn=2+5+(3n-4)+(3n-1),-得Tn=2+3-(3n-1)=1+3-(3n-1),整理得Tn=-(3n+5)+5.导学号324708702.已知数列an的前n项和为Sn,且Sn=2an-1;数列bn满足bn-1-bn=bnbn-1(n2,nN+),b1=1.(1)求数列an,bn的通项公式;(2)求数列的前n项和Tn.解:(1)由Sn=2an-1,得S1=a1=2a1-1,故a1=1.又Sn=2an-1,Sn-1=2an-1-1(n2),两式相减,得Sn-Sn-1=2an-2an-1,即an=2an-2an-1.故an=2an-1,n2.所以数列an是首项为1,公比为2的等比数列.故an=12n-1=2n-1.由bn-1-bn=bnbn-1(n2,nN+),得=1.又b1=1,数列是首项为1,公差为1的等差数列.=1+(n-1)1=n.bn=.(2)由(1)得=n2n-1.Tn=120+221+n2n-1,2Tn=121+222+n2n.两式相减,得-Tn=1+21+2n-1-n2n=-n2n=-1+2n-n2n.Tn=(n-1)2n+1.导学号324708713.(2015山东滨州一模)已知数列an的前n项和是Sn,且Sn+an=1(nN+).(1)求数列an的通项公式;(2)设bn=lo(1-Sn+1)(nN+),令Tn=+,求Tn.解:(1)当n=1时,a1=S1,由S1+a1=1,得a1=.当n2时,Sn=1-an,Sn-1=1-an-1,则Sn-Sn-1=(an-1-an),即an=(an-1-an),所以an=an-1(n2).故数列an是以为首项,为公比的等比数列.故an=2.(2)因为1-Sn=an=,所以bn=lo(1-Sn+1)=lo=n+1,因为,所以Tn=+=+=.导学号324708724.(2015江西上饶一模)已知数列an的前n项和Sn=2an-32n+4(nN+).(1)证明:数列是等差数列;(2)设bn=,求数列bn的前n项和Tn.(1)证明:Sn=2an-32n+4(nN+),n=1时,a1=S1=2a1-6+4,解得a1=2.当n2时,an=Sn-Sn-1=2an-32n+4-(2an-1-32n-1+4),化为an=2an-1+32n-1,变形为,数列是等差数列,首项为=1,公差为.(2)解:由(1)可得=1+(n-1)=,bn=,数列bn的前n项和Tn=+.导学号324708735.(2015长沙二模改编)已知数列an是公差不为零的等差数列,a10=15,且a3,a4,a7成等比数列.(1)求数列an的通项公式;(2)设bn=,求数列bn的前n项和Tn.解:(1)设数列an的公差为d(d0),由已知得即解得an=2n-5.(2)证明:bn=,nN+,Tn=+,Tn=+,-,得Tn=+2+=-,Tn=-1-(nN+).导学号324708746.已知等差数列an的公差大于0,且a3,a5是方程x2-14x+45=0的两根,数列bn的前n项和为Sn,且Sn=(nN+).(1)求数列an,bn的通项公式;(2)记cn=anbn,求证:cn+1cn;(3)求数列cn的前n项和Tn.解:(1)因为a3,a5是方程x2-14x+45=0的两根,且数列an的公差d0,所以a3=5,a5=9,公差d=2.所以an=a5+(n-5)d=2n-1.当n=1时,b1=S1=,解得b1=.当n2时,bn=Sn-Sn-1=(bn-1-bn),所以(n2).所以数列bn是首项b1=,公比q=的等比数列,所以bn=b1qn-1=.(2)由(1),知cn=anbn=,cn+1=,所以cn+1-cn=0.所以cn+1cn.(3)由(2),知cn=anbn=,则Tn=+,Tn=+,-,得Tn=+2,化简得Tn=1-.故数列cn的前n项和Tn=1-.导学号324708752