第2讲数列递推与通项选题明细表知识点·方法巩固提高A巩固提高B利用递推直接求通项2,6,9,131,7,10,11,16已知Sn求an1,8,10,11,149,12,13累加法求通项4,55,8,15,17累乘法求通项7,176,14构造新数列求通项3,12,15,162,3,4巩固提高A一、选择题1.已知Sn为数列{an}的前n项和,若a2=3且Sn+1=2Sn,则a4等于(B)(A)6(B)12(C)16(D)24解析:由S2=2S1=2a1=a1+a2=a1+3得a1=3,S3=2S2=2(a1+a2)=12,S4=2S3=24,所以a4=S4-S3=12,故选B.2.已知数列{an}中,a1=3,a2=5且对于大于2的正整数,总有an=an-1-an-2,则a2018等于(B)(A)-5(B)5(C)-3(D)3解析:an+6=an+5-an+4=an+4-an+3-an+4=-(an+2-an+1)=-an+2+an+1=-(an+1-an)+an+1=an,故数列{an}是以6为周期的周期数列,所以a2018=a336×6+2=a2=5,故选B.3.已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,bn+1=,则{bn}的通项公式bn等于(B)(A)2n-1(B)2n+1(C)2n+1-1(D)2n-1+2解析:易得an=2n-1,故由bn+1=可得bn+1=2bn-1,变形为bn+1-1=2(bn-1),即数列{bn-1}是首项为2,公比为2的等比数列,故bn-1=2n,解得bn=2n+1,故选B.4.数列{an}的首项为3,{bn}为等差数列,且bn=an+1-an(n∈N*),若b3=-2,b10=12,则a8等于(B)(A)0(B)3(C)8(D)11解析:由题意可设等差数列{bn}的首项为b1,公差为d,所以d===2,所以b1=b3-2d=-2-4=-6,所以bn=2n-8,即an+1-an=2n-8,an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=3+(-6)+(-4)+…+(2n-10)=3+(n-8)(n-1),所以a8=3,选B.5.已知数列{an}满足a1=1,且an=an-1+()n(n≥2,且n∈N*),则数列{an}的通项公式为(B)(A)an=(B)an=(C)an=n+2(D)an=(n+2)3n解析:由an=an-1+()n(n≥2且n∈N*),得3nan=3n-1an-1+1,3n-1an-1=3n-2an-2+1,…,32a2=3a1+1,以上各式相加得3nan=n+2,故an=,当n=1时也适合,所以an=.6.已知数列{an}和{bn}满足a1a2…an=((n∈N*).若{an}为等比数列,且a1=2,b3=6+b2,则an与bn分别为(B)(A)an=2n-1,bn=n(n+1)(B)an=2n,bn=n(n+1)(C)an=2n,bn=(n-1)n(D)an=2n+1,bn=(n-1)n解析:设等比数列{an}的公比为q,因为b3=6+b2,即b3-b2=6,所以a3=(=()6=8,所以2q2=8,又由题意得an>0,所以q=2,所以an=2×2n-1=2n.所以a1a2…an=21+2+…+n==(=,所以bn=n(n+1).选B.7.已知数列{an}的前n项和为Sn,且满足4(n+1)(Sn+1)=(n+2)2an,则数列{an}的通项公式an等于(A)(A)(n+1)3(B)(2n+1)2(C)8n2(D)(2n+1)2-1解析:当n=1时,4(1+1)(a1+1)=(1+2)2a1,解得a1=8,当n≥2时,由4(Sn+1)=,得4(Sn-1+1)=,两式相减得,4an=-,即=,所以=··…·=××…×=,所以an=a1×=(n+1)3,经验证n=1时也符合,所以an=(n+1)3.8.已知数列{an}的首项a1=a,其前n项和为Sn,且满足Sn+Sn-1=4n2(n≥2,n∈N+),若对任意n∈N+,an
0,将an+1=两边取对数得lgan+1=2lgan,即=2,所以数列{lgan}是以lga1=lg3为首项,公比为2的等比数列,lgan=lga1·2n-1=lg,即an=.答案:13.数阵(数表)中涉及的数列通项公式问...
