数列经典试题含答案

相交,故每增加一条直线一定与前面已有的每条直线都相交,∴f(k)=f(k-1)+(k-1).由f(3)=2,f(4)=f(3)+3=2+3=5,f(5)=f(4)+4=2+3+4=9,……f(n)=f(n-1)+(n-1),相加得f(n)=2+3+4+…+(n-1)=(n+1)(n-2).三、解答题17.分析:判定给定数列是否为等差数列关键看是否满足从第2项开始每项与其前一项差为常数.证明:(1)n=1时,a1=S1=3-2=1,当n≥2时,an=Sn-Sn-1=3n2-2n-[3(n-1)2-2(n-1)]=6n-5,n=1时,亦满足,∴an=6n-5(n∈N*).首项a1=1,an-an-1=6n-5-[6(n-1)-5]=6(常数)(n∈N*),∴数列{an}成等差数列且a1=1,公差为6.(2)∵,,成等差数列,∴=+化简得2ac=b(a+c).+=====2·,∴,,也成等差数列.18.解:(1)由题设2a3=a1+a2,即2a1q2=a1+a1q,∵a1≠0,∴2q2-q-1=0,∴q=1或-.(2)若q=1,则Sn=2n+=.当n≥2时,Sn-bn=Sn-1=>0,故Sn>bn.若q=-,则Sn=2n+(-)=.当n≥2时,Sn-bn=Sn-1=,故对于n∈N+,当2≤n≤9时,Sn>bn;当n=10时,Sn=bn;当n≥11时,Sn<bn.19.证明:∵an+1=Sn+1-Sn,an+1=Sn,∴(n+2)Sn=n(Sn+1-Sn),整理得nSn+1=2(n+1)Sn,所以=.故{}是以2为公比的等比数列.20.证明:由a1,2a7,3a4成等差数列,得4a7=a1+3a4,即4a1q6=a1+3a1q3,变形得(4q3+1)(q3-1)=0,∴q3=-或q3=1(舍).由===;=-1=-1=1+q6-1=;得=.∴12S3,S6,S12-S6成等比数列.