湘教版高中数学选择性必修第一册专题强化练3数列的递推公式及通项公式含答案

专题强化练3数列的递推公式及通项公式(2020河北唐山一中月考)已知数列{an}中,a1=3,a2=6,an+2=an+1-an,则a2020=()A.6B.-6C.3D.-32.已知数列{an}满足a1=2,an+1=an2,则数列{an}的通项公式为aA.2n-1B.2n-1C.22n3.(2022湖南邵东一中月考)记首项为1的数列{an}的前n项和为Sn,且n≥2时,an(2Sn-1)=2Sn2,则S10A.110B.113C.14.(2022河南许昌期末)数列{an}的首项a1=2,且an+1=4an+6,令bn=log2(an+2),则b1A.2020B.2021C.2022D.20235.在数列{an},{bn}中,an+1=an+bn+an2+bn2,bn+1=an+bn-an2+bn2,a1=1,b1=1,设cn=A.2016B.4020C.2020D.40406.(2022江苏南京十三中期末)已知数列{an}的前n项和为Sn,a1=1,a2=3,且Sn+1+Sn-1=2n+2Sn(n≥2),若λ(Sn-an)+λ+7≥(2-λ)n对任意n∈N+都成立,则实数λ的最小值为()A.-52B.116C.7.已知在数列{an}中,a1=2,a1+a22+a33+…+ann=an+1-2,8.已知数列{an}的前n项和为Sn,且a1=2,Sn+1=4an+2,则an=.

9.[2021新高考八省(市)联考]已知各项都为正数的数列{an}满足an+2=2an+1+3an.(1)证明:数列{an+an+1}为等比数列;(2)若a1=12,a2=32,求{an}

10.(2022广东深圳龙城中学月考)在①nan+1-(n+1)an=n2+n,②3Sn=(n+2)an,③Tn+1=(n+2)an设首项为2的数列{an}的前n项和为Sn,前n项积为Tn,且.

(1)求数列{an}的通项公式;(2)设bn=(-1)nan,求数列{bn}的前n项和.

答案与分层梯度式解析1.D因为an+2=an+1-an①,所以an+3=an+2-an+1②,①+②并化简得an+3=-an,则an+6=-an+3=an,即数列{an}是周期为6的周期数列.因为a1=3,a2=6,所以a3=a2-a1=3,a4=-a1=-3,a5=-a2=-6,a6=-a3=-3,则a2020=a6×336+4=a4=-3,故选D.2.C易知an>0,且an≠1,在an+1=an2的两边同时取常用对数,得lgan+1=2lgan,故lgan+1lgan=2,所以数列{lgan}是以lg2为首项,2为公比的等比数列,所以lgan=2n-1×lg2=lg223.D当n≥2时,an(2Sn-1)=2Sn2,则(Sn-Sn-1)(2Sn-1)=2Sn2,即Sn+2SnSn-1=Sn-1,可得所以1Sn是首项为1、公差为2所以1Sn=1+2(n-1)=2n-1,所以Sn=所以S10=12×10-1=119,4.C∵an+1=4an+6,∴an+1+2=4(an+2)>0,即an+1+2an+2=4,又∵a1=2,∴数列{an+2}是以4为首项、4为公比的等比数列,故an+2=4n.由bn=log2(an+2),得bn=log24n=2n,设数列{bn}的前n项和为Sn,则S2021=2×(1+2+3+…+2020+2∴b1+b2+…+5.Dan+1=an+bn+an2+bn2①,bn+1=an+bn-an2+bn2②,①+②,得an+1+bn+1=2(an+bn),又因为a1=1,b1=1,所以{an+bn}是以2为首项、2为公比的等比数列,所以an+bn=2n.①×②,得an+1bn+1=2anbn,又因为a1=1,b1=1,所以{anbn}是以1为首项、2为公比的等比数列,所以anbn=2n-1.所以cn=an+bna6.C由题意得Sn+1-Sn=2n+Sn-Sn-1(n≥2),故an+1-an=2n(n≥2),因为a2-a1=2,所以an+1-an=2n(n≥1),所以an-an-1=2n-1,an-1-an-2=2n-2,……,a2-a1=2,则an-a1=21+22+…+2n-1,故an=1+21+22+…+2n-1=1-2n1-2=2n-1,所以Sn=21+22+23+…+2n-n=2(1-所以Sn-an=2n-n-1.因为λ(Sn-an)+λ+7≥(2-λ)·n对任意n∈N+都成立,所以λ≥2n设cn=2n-72n,则cn+1-cn=2n-52n+1-2n-72n=9-2n2n+1,当n≤4时,cn+1>cn,当n≥5时,cn+1<cn,因此c1<c2<c3<c4<c5>c6>c7>7.答案2n解析因为a1+a22+a33+…+ann=an+1-2①,所以当n=1时,a1=a2-2,得a2=4;当n≥2时,a1+a22+a33+…+an-1n-1=an即nan+1=(n+1)·an,易知an≠0,所以an+1an=因为a2a1=2,所以an+1an=n+1n(n≥1),所以a2a1·a3a2·a4a3·…·a8.答案n·2n解析由题意得S2=4a1+2,所以a1+a2=4a1+2,解得a2=8,又因为an+2=Sn+2-Sn+1=4an+1-4an,于是an+2-2an+1=2(an+1-2an),因此数列{an+1-2an}是以a2-2a1=4为首项、2为公比的等比数列,故an+1-2an=4×2n-1=2n+1,于是an+12n+1-an2n=1,因此数列an2n是以1为首项、1为公差的等差数列,9.解析(1)证明:∵an+2=2an+1+3an,∴an+2+an+1=3(an+1+an),易知an+an+1≠0,∴{an+an+1}是以a1+a2为首项、3为公比的等比数列.(2)由题意得a1+a2=2,∴an+1+an=2·3n-1,∴an=2·3n-2-an-1=2·3n-2-(2·3n-3-an-2)=4·3n-3+an-2(n≥3).当n≥3且n为奇数时,an=4·3n-3+4·3n-5+…+4·30+a1=4·1-9n-121-9+12=12·3n-1,当n=1当n≥3且n为偶数时,an=4·3n-3+4·3n-5+…+4·31+a2=4·3(1-9n-22)1-9+32=12·3n-1,当综上,an=12·3n-1(n∈N+10.解析(1)选①.因为nan+1-(n+1)an=n2+n,所以an+1n因为a1=2,所以a11=2,所以数列ann是以2为首项、1为公差的等差数列,所以ann=n+1,选②.因为3Sn=(n+2)an,所以3Sn+1=(n+3)an+1,两式相减,可得3an+1=(n+3)an+1-(n+2)an,即nan+1=(n+2)an,所以an+1(n+2)(n+1)=an(n+1)n,所以an(选③.因为Tn+1=(n+2)anTnn,所以Tn+1Tn=an+1=(n+2)ann,即an+1((2)由(1)可知,an=n2+n,所以bn=(-1)n(n2+