限时集训（十一）数列求和及数列的简单应用基础过关1.已知各项均为正数且递减的等比数列{an}满足a3,32a4,2a5成等差数列,其前5项和S5=31.(1)求数列{an}的通项公式;(2)若等差数列{bn}满足b1=a4-1,b2=a3-1,求数列{abn}的前n项和Tn.2.已知数列{an}满足a1=a3,an+1-an2=32n+1,设bn=2nan.(1)求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.3.已知数列{an}为正项数列,a1=4,且对任意n∈N*,an+12-2an2=anan+1恒成立.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=1n·log2an,Tn为数列{bn}的前n项和,求证:Tn<1.4.已知{an}是等比数列,数列{bn}满足b1=-2,b2=5,且a1b1+a2b2+…+anbn=2+(2n-3)·4n.(1)求{an}的通项公式和前n项和Sn;(2)求{bn}的通项公式.能力提升5.已知数列{an}满足an+1+1=an+1an+2,an≠-1且a1=1.(1)证明数列{1an+1}是等差数列,并求出数列{an}的通项公式;(2)令bn=2nan+1,求数列{bn}的前n项和Sn.6.已知数列{an},{bn},其中a1=3,b1=-1,且满足an=12(3an-1-bn-1),bn=-12(an-1-3bn-1),n∈N*,n≥2.(1)求证:数列{an-bn}为等比数列;(2)求数列{2nanan+1}的前n项和Sn.限时集训(十一)基础过关1.解:(1)设{an}的公比为q(00,∴an+1=2an,∴数列{an}是首项为4,公比为2的等比数列,∴an=2n+1.(2)证明:bn=1n·log2an=1n(n+1¿¿=1n-1n+1,∴Tn=b1+b2+…+bn=11-12+12-13+…+1n-1n+1=1-1n+1<1.4.解:(1)∵a1b1+a2b2+…+anbn=2+(2n-3)·4n,∴a1b1=2-4=-2,a1b1+a2b2=2+(4-3)×42=18,则a2b2=20,又b1=-2,b2=5,∴a1=1,a2=4,∵{an}是等比数列,a2a1=4,∴{an}的通项公式为an=4n-1,∴{an}的前n项和Sn=1−4n1−4=4n-13.(2)由an=4n-1及a1b1+a2b2+…+anbn=2+(2n-3)·4n,得b1+4b2+…+4n-1bn=2+(2n-3)·4n,①当n≥2时,b1+4b2+…+4n-2bn-1=2+(2n-5)·4n-1,②①-②得4n-1bn=2+(2n-3)·4n-2-(2n-5)·4n-1=(6n-7)·4n-1,∴bn=6n-7.又当n=1时,b1=-2,不满足上式,∴{bn}的通项公式为bn={-2,n=1,6n-7,n≥2.能力提升5.解:(1)∵an+1+1=an+1an+2,an≠-1且a1=1,∴1an+1+1=an+2an+1,即1an+1+1=(an+1¿+1¿an+1,∴1an+1+1-1an+1=1,∴数列{1an+1}是等差数列.∵1a1+1=12,∴1an+1=12+(n-1)·1,∴1an+1=2n-12,∴an=3−2n2n-1.(2)由(1)知bn=(2n-1)·2n-1,则Sn=1×20+3×21+5×22+…+(2n-1)×2n-1,2Sn=1×21+3×22+…+(2n-3)×2n-1+(2n-1)×2n,∴-Sn=1+2×2+2×22+…+2×2n-1-(2n-1)×2n=1+2×2¿¿-(2n-1)×2n,∴Sn=-1+22-2n+1+(2n-1)×2n=3-2n+1+(2n-1)×2n=(2n-3)×2n+3.6.解:(1)证明:an-bn=12(3an-1-bn-1)-(-12)(an-1-3bn-1)=2(an-1-bn-1),又a1-b1=3-(-1)=4,所以{an-bn}是首项为4,公比为2的等比数列.(2)由(1)知,an-bn=2n+1.①因为an+bn=12(3an-1-bn-1)+(-12)(an-1-3bn-1)=an-1+bn-1,a1+b1=3+(-1)=2,所以{an+bn}为常数列且an+bn=2,②联立①②得an=2n+1,故2nanan+1=2n(2n+1¿(2n+1+1)¿=12n+1-12n+1+1,所以Sn=(121+1-122+1)+122+1-123+1+…+(12n+1-12n+1+1)=121+1-12n+1+1=13-12n+1+1.
