设数列{an}的前n项和为Sn=2n2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.
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设数列{an}的前n项和为Sn=2n2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)设cn=
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)设cn=
a
(1):当n=1时,a1=S1=2;当n≥2时,an=Sn-Sn-1=2n2-2(n-1)2=4n-2,
故{an}的通项公式为an=4n-2,即{an}是a1=2,公差d=4的等差数列. 设{bn}的公比为q,则b1qd=b1,d=4,∴q= 1 4. 故bn=b1qn-1=2× 1 4n-1,即{bn}的通项公式为bn= 2 4n-1. (II)∵cn= an bn= 4n-2 2 4n-1=(2n-1)4n-1, Tn=c1+c2+…+cn Tn=1+3×41+5×42+…+(2n-1)4n-1 4Tn=1×4+3×42+5×43+…+(2n-3)4n-1+(2n-1)4n 两式相减得,3Tn=-1-2(41+42+43+…+4n-1)+(2n-1)4n= 1 3[(6n-5)4n+5] ∴Tn= 1 9[(6n-5)4n+5]
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