解答必刷卷(三)数列考查范围:第28讲~第32讲题组一真题集训1.[2018·全国卷Ⅲ]等比数列{an}中,a1=1,a5=4a3.(1)求{an}的通项公式;(2)记Sn为{an}的前n项和,若Sm=63,求m.2.[2017·全国卷Ⅲ]设数列{an}满足a1+3a2+…+(2n-1)an=2n.(1)求{an}的通项公式;(2)求数列{an2n+1}的前n项和.3.[2018·天津卷]设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.题组二模拟强化4.[2018·重庆八中月考]已知数列{an}满足a1=1,an-an-1=2n-1(n≥2,n∈N*).(1)求数列{an}的通项公式;(2)设数列bn=log2(an+1),求数列{1bn·bn+1}的前n项和Sn.5.[2018·长春二模]已知数列{an}的通项公式为an=2n-11.(1)求证:数列{an}是等差数列;(2)令bn=|an|,求数列{bn}的前10项和S10.6.[2018·吉林梅河口五中月考]在数列{an}中,a1=1,an+1={13an+n,n,为奇数an-3n,n.为偶数(1)证明:数列a2n-32是等比数列;(2)若Sn是数列{an}的前n项和,求S2n.7.[2018·江西九校二联]已知数列{an}为等差数列,且a2+a3=8,a5=3a2.(1)求数列{an}的通项公式;(2)记bn=2anan+1,设{bn}的前n项和为Sn,求使得Sn>20172018的最小的正整数n.解答必刷卷(三)1.解:(1)设{an}的公比为q,由题设得an=qn-1.由已知得q4=4q2,解得q=0(舍去)或q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,则Sn=1−¿¿.由Sm=63得(-2)m=-188,此方程没有正整数解.若an=2n-1,则Sn=2n-1.由Sm=63得2m=64,解得m=6.综上,m=6.2.解:(1)因为a1+3a2+…+(2n-1)an=2n,故当n≥2时,a1+3a2+…+(2n-3)an-1=2(n-1).两式相减得(2n-1)an=2,所以an=22n-1(n≥2).又由题设可得a1=2,从而{an}的通项公式为an=22n-1.(2)记{an2n+1}的前n项和为Sn,由(1)知an2n+1=2(2n+1¿¿=12n-1-12n+1,则Sn=11-13+13-15+…+12n-1-12n+1=2n2n+1.3.解:(1)设等比数列{bn}的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q>0,所以可得q=2,故bn=2n-1.所以Tn=1−2n1−2=2n-1.设等差数列{an}的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n,所以Sn=n(n+1¿¿2.(2)由(1),有T1+T2+…+Tn=(21+22+…+2n)-n=2׿¿-n=2n+1-n-2.由Sn+(T1+T2+…+Tn)=an+4bn,可得n(n+1¿¿2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍)或n=4.所以,n的值为4.4.解:(1)∵an-an-1=2n-1(n≥2,n∈N*),∴an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a2-a1)+a1,即an=2n-1+2n-2+2n-3+…+22+21+1,则an=1׿¿=2n-1.(2)bn=log2(an+1)=n,则1bn·bn+1=1n(n+1¿¿=1n-1n+1,∴Sn=11-12+12-13+13-14+…+1n-1n+1=1-1n+1=nn+1.5.解:(1)证明:∵an=2n-11,∴an+1-an=2(n+1)-11-2n+11=2(n∈N*),∴数列{an}为等差数列.(2)由(1)得bn=|an|=|2n-11|,∴当n≤5时,bn=|2n-11|=11-2n,当n≥6时,bn=|2n-11|=2n-11.∴S10=[55-2×(1+2+3+4+5)]+[2×(6+7+8+9+10)-55]=50.6.解:(1)证明:设bn=a2n-32,则b1=a2-32=13a1+1-32=-16,因为bn+1bn=a2(n+1)-32a2n-32=13a2n+1+(2n+1)−32a2n-32=13(a2n-6n)+(2n+1¿−32¿a2n-32=13a2n-12a2n-32=13,所以数列a2n-32是以-16为首项,13为公比的等比数列.(2)由(1)得bn=a2n-32=-16·(13)n-1=-12·(13)n,即a2n=-12·(13)n+32,由a2n=13a2n-1+(2n-1),得a2n-1=3a2n-3(2n-1)=-12·(13)n-1-6n+152,所以a2n-1+a2n=-12·(13)n-1+(13)n-6n+9=-2·(13)n-6n+9,故S2n=(a1+a2)+(a3+a4)+…+(a2n-1+a2n)=-2×13+(13)2+…+(13)n-6×(1+2+…+n)+9n=-2×13[1−(13)n]1−13-6·n(n+1¿¿2+9n=(13)n-1-3n2+6n=(13)n-3(n-1)2+2.7.解:(1)设等差数列{an}的公差为d,依题意有{2a1+3d=8,a1+4d=3a1+3d,解得{a1=1,d=2,从而数列{an}的通项公式为an=2n-1,n∈N*.(2)因为bn=2anan+1=12n-1-12n+1,所以Sn=11-13+13-15+…+12n-1-12n+1=1-12n+1.令1-12n+1>20172018,解得n>1008.5,故使得Sn>20172018的最小正整数为1009.
