大题专项练(二)数列A组基础通关1.已知等差数列{an}满足a3-a2=3,a2+a4=14.(1)求{an}的通项公式;(2)设Sn是等比数列{bn}的前n项和,若b2=a2,b4=a6,求S7.解(1)设等差数列{an}的公差为d, a3-a2=3,a2+a4=14.∴d=3,2a1+4d=14,解得a1=1,d=3,∴an=1+3(n-1)=3n-2.(2)设等比数列{bn}的公比为q,b2=a2=4=b1q,b4=a6=16=b1q3,联立解得{b1=2,q=2,或{b1=-2,q=-2.∴S7=2×(27-1)2-1=254,或S7=-2×[1-(-2)7]1-(-2)=-86.2.已知数列{an}的前n项和为Sn,满足a2=15,Sn+1=Sn+3an+6.(1)证明:{an+3}是等比数列;(2)求数列{an}的通项公式以及前n项和Sn.(1)证明在Sn+1=Sn+3an+6中,令n=1,得S2=S1+3a1+6,得a1+a2=a1+3a1+6,即a1+15=4a1+6,解得a1=3.因为Sn+1=Sn+3an+6,所以an+1=3an+6.所以an+1+3an+3=3an+9an+3=3.所以{an+3}是以6为首项,3为公比的等比数列.(2)解由(1)得an+3=6×3n-1=2×3n,所以an=2×3n-3.∴Sn=2×(3+32+33+…3n)-3n=2×3×(1-3n)1-3-3n=3n+1-3-3n.3.设数列{an}的前n项和为Sn,Sn=1-an(n∈N*).(1)求数列{an}的通项公式;(2)设bn=log2an,求数列{1bnbn+1}的前n项和Tn.解(1)因为Sn=1-an(n∈N*),所以Sn-1=1-an-1(n∈N*,且n≥2),则Sn-Sn-1=(1-an)-(1-an-1)(n∈N*,且n≥2).即an=12an-1(n∈N*,且n≥2).因为Sn=1-an(n∈N*),所以S1=1-a1=a1,即a1=12.所以{an}是以12为首项,12为公比的等比数列.故an=(12)n(n∈N*).(2)bn=log2an,所以bn=log2(12)n=-n.所以1bnbn+1=1n(n+1)=1n−1n+1,故Tn=(1-12)+(12-13)+…+(1n-1n+1)=1-1n+1=nn+1.4.设等差数列{an}的公差为d,d为整数,前n项和为Sn,等比数列{bn}的公比为q,已知a1=b1,b2=2,d=q,S10=100,n∈N*.(1)求数列{an}与{bn}的通项公式;(2)设cn=anbn,求数列{cn}的前n项和Tn.解(1)由题意可得{10a1+45d=100,a1d=2,解得{a1=9,d=29(舍去)或{a1=1,d=2,所以an=2n-1,bn=2n-1.(2) cn=anbn,cn=2n-12n-1,∴Tn=1+32+522+723+…+2n-12n-1,①12Tn=12+322+523+724+925+…+2n-12n,②①-②可得12Tn=2+12+122+…+12n-2−2n-12n=3-2n+32n,故Tn=6-2n+32n-1.5.已知正项数列{an}的前n项和为Sn,满足2Sn+1=2an2+an(n∈N*).(1)求数列{an}的通项公式;(2)已知对于n∈N*,不等式1S1+1S2+1S3+…+1Sn0,所以a1=1,当n≥2时,2Sn+1=2an2+an(n∈N*),2Sn-1+1=2an-12+an-1(n∈N*),作差整理,得an+an-1=2(an+an-1)(an-an-1),因为an>0,故an+an-1>0,所以an-an-1=12,故数列{an}为等差数列,所以an=n+12.(2)由(1)知Sn=n(n+3)4,所以1Sn=4n(n+3)=43(1n-1n+3),从而1S1+1S2+1S3+…+1Sn=43(1-14)+(12-15)+(13-16)+…+(1n-2-1n+1)+(1n-1-1n+2)+(1n-1n+3)=431+12+13−1n+1−1n+2−1n+3=43116−1n+1−1n+2−1n+3<229.所以M≥229,故M的最小值为229.6.已知数列{an}是公比为q的正项等比数列,{bn}是公差d为负数的等差数列,满足1a2−1a3=da1,b1+b2+b3=21,b1b2b3=315.(1)求数列{an}的公比q与数列{bn}的通项公式;(2)求数列{|bn|}的前10项和S10.解(1)由已知,b1+b2+b3=3b2=21,得b2=7,又b1b2b3=(b2-d)·b2·(b2+d)=(7-d)·7·(7+d)=343-7d2=315,得d=-2或2(舍),b1=7+2=9,bn=-2n+11.于是1a2−1a3=-2a1,又{an}是公比为q的等比数列,故1a1q−1a1q2=-2a1,所以,2q2+q-1=0,q=-1(舍)或12.综上,q=12,d=-2,bn=11-2n.(2)设{bn}的前n项和为Tn;令bn≥0,11-2n≥0,得n≤5,于是,S5=T5=5(b1+b5)2=25.易知,n>6时,bn<0,|b6|+|b7|+…+|b10|=-b6-b7-…-b10=-(b6+b7+…+b10)=-(T10-T5)=-(0-25)=25,所以,S10=50.B组能力提升7.已知数列{an}的前n项和为Sn,点(n,Sn)(n∈N*)在函数f(x)=12x2+12x的图象上.(1)求数列{an}的通项公式;(2)设数列{1anan+2}的前n项和为Tn,不等式Tn>13loga(1-a)对任意正整数n恒成立,求实数a的取值范围.解(1) 点(n,Sn)在函数f(x)=12x2+12x的图象上,∴Sn=12n2+12n.①当n≥2时,Sn-1=12(n-1)2+12(n-1),②①-②,得an=n.当n=1时,a1=S1=1,符合上式.∴an=n(n∈N*).(2)由(1),得1anan+2=1n(n+2)=12(1n-1n+2),∴Tn=1a1a3+1a2a4+…+1anan+2=121-13+12−14+…+1n−1n+2=34−121n+1+1n+2. Tn+1-Tn=1(n+1)(n+3)>0,∴数列{Tn}单调递增,∴{Tn}中的最小项为T1=13.要使不等式Tn>13loga(1-a)对任意正整数n恒成立,...
