2019届高考数学总复习模块三数列限时集训十一数列求和及数列的简单应用理.docx

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nN*,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=1nlog2an,Tn为数列bn的前n项和,求证:Tn1.4.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),nN*,n2.(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(0q0,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=1nlog2an=1n(n+1)=1n-1n+1,Tn=b1+b2+bn=11-12+12-13+1n-1n+1=1-1n+11.4.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,当n2时,b1+4b2+4n-2bn-1=2+(2n-5)4n-1,-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,n2. 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)+1an+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=120+321+522+(2n-1)2n-1,2Sn=121+322+(2n-3)2n-1+(2n-1)2n,-Sn=1+22+222+22n-1-(2n-1)2n=1+22(1-2n-1)1-2-(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.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,联立得an=2n+1,故2nanan+1=2n(2n+1)(2n+