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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?,知xn?1?2???2?(Ⅱ)由xn?1?,同理xn?1?2?. 2xn2xn2xn2xn 故
xn?1?2x?22?(n).
xn?1?2xn?2xn?1?2x?2?2lgn,即an?1?2an.所以,数列{an}成等比数列.
xn?1?2xn?2n?1从而lg故an?2a1?2n?1lgx1?2?2n?1lg3. x1?2即lgxn?2?2n?1lg3. xn?2n?1xn?2?32 xn?2从而
所以xn?2(32?1)32n?1n?1?12(3
2n?1n?1(Ⅲ)由(Ⅱ)知xn??1)32?1,
∴bn?xn?2?n?1432n?1?1?0
bn?132?11111∴?2n?2n?1?2n?1?21?1?
bn33?13?133当n?1时,显然T1?b1?2?3. 当n?1时,bn?111bn?1?()2bn?2?L?()n?1b1 333∴Tn?b1?b2?L?bn
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?b13?L?(11?b13)n?1b1
b[1?(1?1)n3] 1?13?3?3?(13)n?3.
综上,Tn?3(n?N*). / 14
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