2即y?(xn?4)?2xn(x?xn)．
2令y?0，得?(xn?4)?2xn(xn?1?xn)． 2即xn?4?2xnxn?1．
显然xn?0，∴xn?1?xn2?． 2xnxn2xn2(xn?2)2(xn?2)2（Ⅱ）由xn?1?，同理xn?1?2?． ?，知xn?1?2???2?2xn2xn2xn2xnx?2x?22x?2x?2 故n?1，即an?1?2an．所以，数列?(n)．从而lgn?1?2lgnxn?1?2xn?2xn?1?2xn?2x?2x?2?2n?1lg3．即lgn?2n?1lg3． {an}成等比数列．故an?2n?1a1?2n?1lg1x1?2xn?2xn?22(32?1)2n?1从而 ?3所以xn?2n?1xn?23?1（Ⅲ）由（Ⅱ）知xn?n?12(32?1)32n?1n?1?1n?14bn?132?11111?0∴∴bn?xn?2?2n?1?2n?2n?1?2n?1?21?1?
bn33?13?13?1331121n?1当n?1时，显然T1?b1?2?3．当n?1时，bn?bn?1?()bn?2???()b1
3331b1[1?()n]11n?113?3?3?()n?3． ∴Tn?b1?b2???bn?b1?b1???()b1?13331?3 综上，Tn?3(n?N*)．
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