限时集训（十一）数列求和及综合应用基础过关1.已知数列{an}是首项为a1,公比为q的等比数列,Sn为其前n项和.(1)若a2=2,且a3是S1,S3的等差中项,求数列{an}的通项公式;(2)当a1=1,q=2时,令bn=log4(Sn+1),求证:数列{bn}是等差数列.2.已知数列{an}满足a1+13a2+15a3+…+12n-1an=n(n∈N*).(1)求数列{an}的通项公式;(2)求数列2❑√an+1+❑√an的前60项和T60.3.已知数列{an}是等差数列,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,求最小的正整数n,使得Sn>20172018.4.已知数列{an}满足a1=12,an+1=an2an+1(n∈N*).(1)证明数列1an是等差数列,并求{an}的通项公式;(2)若数列{bn}满足bn=12n·an,求数列{bn}的前n项和Sn.能力提升5.在公比为q的等比数列{an}中,已知a1=16,且a1,a2+2,a3成等差数列.(1)求q,an;(2)若q<1,求使得a1-a2+a3-…+(-1)2n-1a2n>10的正整数n的最小值.6.已知数列{an}是递增的等比数列,a1=1,a3=4,数列{bn}满足b1=2,bn+1-2bn=8an(n∈N*).(1)求数列{an}和{bn}的通项公式;(2)设数列{cn}满足cn=4nbn·bn+1,且数列{cn}的前n项和为Tn,求使得Tn>1am对任意n∈N*都成立的正整数m的最小值.限时集训(十一)基础过关1.解:(1)由题意得{a2=2,2a3=S1+S3,从而有{a1q=2,2a1q2=a1+a1+a1q+a1q2,解得{q=2,a1=1或{q=−1,a1=−2,所以an=2n-1或an=2×(-1)n.(2)证明:由题意得Sn=2n-1,所以bn=log4(Sn+1)=n2.当n≥2时,因为bn-bn-1=n2-n-12=12,所以数列{bn}是公差为12的等差数列.2.解:(1)∵数列{an}满足a1+13a2+15a3+…+12n-3an-1+12n-1an=n①,∴当n≥2时,a1+13a2+15a3+…+12n-3an-1=n-1②,由①-②得12n-1an=1,所以an=2n-1.当n=1时,a1=1适合上式,∴an=2n-1.(2)令bn=2❑√an+1+❑√an,则bn=2❑√2n+1+❑√2n-1=❑√2n+1-❑√2n-1,∴Tn=(❑√3-❑√1)+(❑√5-❑√3)+…+(❑√2n+1-❑√2n-1)=❑√2n+1-1,∴T60=10.3.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.(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,又n∈N*,故取n=1009,所以使得Sn>20172018的最小的正整数n=1009.4.解:(1)证明:∵an+1=an2an+1,∴1an+1-1an=2,∴数列1an是首项为1a1=2,公差为2的等差数列.∴1an=2+2(n-1)=2n,∴an=12n.(2)∵bn=2n2n=n2n-1,∴Sn=b1+b2+…+bn=1+22+322+…+n2n-1,12Sn=12+222+323+…+n2n,两式相减得12Sn=1+12+122+123+…+12n-1-n2n=21-12n-n2n=2-2+n2n,∴Sn=4-2+n2n-1.能力提升5.解:(1)由题意得2(a2+2)=a1+a3,即2(16q+2)=16+16q2,整理得4q2-8q+3=0,解得q=12或q=32.当q=12时,an=25-n;当q=32时,an=16×32n-1.(2)由(1)知若q<1,则q=12,an=25-n,数列a1,-a2,a3,-a4,…,(-1)2n-1a2n是首项为a1=16,公比为-12的等比数列,则a1-a2+a3-…+(-1)2n-1a2n=16[1−(−12)2n]1−(−12)=3231-(−12)2n,由3231-(-12)2n>10,得122n<116,即2n>4,n>2,∴使得a1-a2+a3-…+(-1)2n-1a2n>10的正整数n的最小值为3.6.解:(1)∵数列{an}是递增的等比数列,∴其公比q>1,又q2=a3a1=4,∴q=2,∴an=2n-1,∴bn+1-2bn=2n+2,∴bn+12n+1-bn2n=2,∴数列{bn2n}是首项为b12=1,公差为2的等差数列,∴bn2n=1+2(n-1)=2n-1,∴bn=2n(2n-1).(2)∵cn=4nbn·bn+1=12(2n+1)¿¿=1412n-1-12n+1,∴Tn=141-13+13-15+…+12n-1-12n+1=141-12n+1,显然数列{Tn}是递增数列,∴当n=1时,Tn取得最小值16,要使得Tn>1am对任意n∈N*都成立,只需16>1am,即16>12m-1,又∵m∈N*,∴m≥4,故正整数m的最小值为4.
