摘要:解:设数列{an}的公比为q.则q2==4.由an>0(n∈N*).得q=2.∴a1=1.an=2n.
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已知数列{an}的首项a1=1,前n项之和Sn满足关系式:3tSn+1-(2t+3)Sn=3t(t>0,n∈N*).
(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足bn+1=f(
),(n∈N*),且b1=1.
(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.
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(1)求证:数列{an}是等比数列;
(2)设数列{an}的公比为f(t),数列{bn}满足bn+1=f(
| 1 | bn |
(i)求数列{bn}的通项bn;
(ii)设Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1,求Tn.