4.2数列解答题命题角度1等差、等比数列的判定与证明高考真题体验·对方向1.(2019全国Ⅱ·19)已知数列{an}和{bn}满足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)证明:{an+bn}是等比数列,{an-bn}是等差数列;(2)求{an}和{bn}的通项公式.(1)证明由题设得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=12(an+bn).又因为a1+b1=1,所以{an+bn}是首项为1,公比为12的等比数列.由题设得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因为a1-b1=1,所以{an-bn}是首项为1,公差为2的等差数列.(2)解由(1)知,an+bn=12n-1,an-bn=2n-1.所以an=12[(an+bn)+(an-bn)]=12n+n-12,bn=12[(an+bn)-(an-bn)]=12n-n+12.2.(2014全国Ⅱ·17)已知数列{an}满足a1=1,an+1=3an+1.(1)证明{an+12}是等比数列,并求{an}的通项公式;(2)证明1a1+1a2+…+1an<32.解(1)由an+1=3an+1得an+1+12=3(an+12).又a1+12=32,所以{an+12}是首项为32,公比为3的等比数列.an+12=3n2,因此{an}的通项公式为an=3n-12.(2)由(1)知1an=23n-1.因为当n≥1时,3n-1≥2×3n-1,所以13n-1≤12×3n-1.于是1a1+1a2+…+1an≤1+13+…+13n-1=32(1-13n)<32.所以1a1+1a2+…+1an<32.典题演练提能·刷高分1.(2019江西临川第一中学高三下学期考前模拟)已知数列{an}中,a1=m,且an+1=3an+2n-1,bn=an+n(n∈N*).(1)判断数列{bn}是否为等比数列,并说明理由;(2)当m=2时,求数列{(-1)nan}的前2020项和S2020.解(1) an+1=3an+2n-1,∴bn+1=an+1+n+1=3an+2n-1+n+1=3(an+n)=3bn.①当m=-1时,b1=0,数列{bn}不是等比数列;②当m≠-1时,数列{bn}是等比数列,其首项为b1=m+1≠0,公比为3.(2)由(1)且当m≠-1时,有bn=an+n=3×3n-1=3n,即an=3n-n,∴(-1)nan=(-3)n-(-1)nn.∴S2020=-3×[1-(-3)2020]1-(-3)-[(-1+2)+(-3+4)+…+(-2019+2020)]=-3+320214-1010=32021-40434.2.(2019重庆一中高三下学期5月月考)已知数列{an}满足:an≠1,an+1=2-1an(n∈N*),数列{bn}中,bn=1an-1,且b1,b2,b4成等比数列.(1)求证:数列{bn}是等差数列;(2)若Sn是数列{bn}的前n项和,求数列{1Sn}的前n项和Tn.(1)证明bn+1-bn=1an+1-1−1an-1=12-1an-1−1an-1=anan-1−1an-1=1,所以数列{bn}是公差为1的等差数列.(2)解由题意可得b22=b1b4,即(b1+1)2=b1(b1+3),∴b1=1.故bn=n.所以Sn=n(n+1)2.所以1Sn=2n(n+1)=21n−1n+1.所以Tn=2×1-12+12−13+…+1n−1n+1=2×1-1n+1=2nn+1.3.已知数列{an}的前n项和为Sn,且Sn=2an-1.(1)证明数列{an}是等比数列;(2)设bn=(2n-1)an,求数列{bn}的前n项和Tn.解(1)当n=1时,a1=S1=2a1-1,所以a1=1,当n≥2时,an=Sn-Sn-1=(2an-1)-(2an-1-1),所以an=2an-1,所以数列{an}是以a1=1为首项,以2为公比的等比数列.(2)由(1)知,an=2n-1,所以bn=(2n-1)2n-1,所以Tn=1+3×2+5×22+…+(2n-3)·2n-2+(2n-1)·2n-1①,2Tn=1×2+3×22+…+(2n-3)·2n-1+(2n-1)·2n②,①-②得-Tn=1+2(21+22+…+2n-1)-(2n-1)·2n=1+2×2-2n-1×21-2-(2n-1)2n=(3-2n)2n-3,所以Tn=(2n-3)2n+3.4.设a1=2,a2=4,数列{bn}满足:bn+1=2bn+2且an+1-an=bn.(1)求证:数列{bn+2}是等比数列;(2)求数列{an}的通项公式.解(1)由题知bn+1+2bn+2=2bn+2+2bn+2=2,又 b1=a2-a1=4-2=2,∴b1+2=4,∴{bn+2}是以4为首项,以2为公比的等比数列.(2)由(1)可得bn+2=4×2n-1,故bn=2n+1-2. an+1-an=bn,∴a2-a1=b1,a3-a2=b2,a4-a3=b3,…,an-an-1=bn-1.累加得an-a1=b1+b2+b3+…+bn-1,an=2+(22-2)+(23-2)+(24-2)+…+(2n-2)=2+22(1-2n-1)1-2-2(n-1)=2n+1-2n,即an=2n+1-2n(n≥2).而a1=2=21+1-2×1,∴an=2n+1-2n(n∈N*).命题角度2等差、等比数列的通项公式与前n项和公式的应用高考真题体验·对方向1.(2019天津·19)设{an}是等差数列,{bn}是等比数列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求{an}和{bn}的通项公式;(2)设数列{cn}满足c1=1,cn={1,2k
