由f(3)=2,Рf(4)=f(3)+3=2+3=5,Рf(5)=f(4)+4=2+3+4=9,Р……Рf(n)=f(n-1)+(n-1),Р相加得f(n)=2+3+4+…+(n-1)=(n+1)(n-2).Р三、解答题Р17.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.Р证明:(1)n=1时,a1=S1=3-2=1,Р当n≥2时,an=Sn-Sn-1=3n2-2n-[3(n-1)2-2(n-1)]=6n-5,Рn=1时,亦满足,∴an=6n-5(n∈N*).Р首项a1=1,an-an-1=6n-5-[6(n-1)-5]=6(常数)(n∈N*),Р∴数列{an}成等差数列且a1=1,公差为6.Р(2)∵,,成等差数列,Р ∴=+化简得2ac=b(a+c).Р +=====2·,Р∴,,也成等差数列.Р18.解:(1)由题设2a3=a1+a2,即2a1q2=a1+a1q,Р∵a1≠0,∴2q2-q-1=0,Р∴q=1或-.Р(2)若q=1,则Sn=2n+=.Р当n≥2时,Sn-bn=Sn-1=>0,故Sn>bn.Р若q=-,则Sn=2n+ (-)=.Р当n≥2时,Sn-bn=Sn-1=,Р故对于n∈N+,当2≤n≤9时,Sn>bn;当n=10时,Sn=bn;当n≥11时,Sn<bn.Р19.证明:∵an+1=Sn+1-Sn,an+1=Sn,Р∴(n+2)Sn=n(Sn+1-Sn),整理得nSn+1=2(n+1) Sn,Р所以=.Р故{}是以2为公比的等比数列.Р20.证明:由a1,2a7,3a4成等差数列,得4a7=a1+3a4,即4 a1q6=a1+3a1q3,Р 变形得(4q3+1)(q3-1)=0,Р ∴q3=-或q3=1(舍).Р 由===;Р =-1=-1=1+q6-1=;Р 得=.Р ∴12S3,S6,S12-S6成等比数列.
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