整理得，(an?an?1)(an?an?1?1)?0. 又
an?0得an?an?1?1，且a1?1，
n所以?an?是首项为1，公差为1的等差数列，即an?n，bn?2.
?nbn?n?2n. ??????????????????????????????7分 Tn?1?21?2?22?3?23?????(n?1)?2n?1?n?2n， 2Tn?1?22?2?23?3?24?????(n?1)?2n?n?2n?1，
由上两式相减得 ?Tn?2?2?2?????2?n?2123nn?12(1?2n)??n?2n?1.
1?2?Tn?(n?1)2n?1?2. ??????????????????????????10分
(3)由(2)知bn?2，只需证ln(1?2n)?2n.设f(x)?ln(1?2x)?2x(x?1且x?R).
n2xln22xln2x则f'(x)??2ln2??(?2x)?0， xx1?21?2可知f(x)在[1,??)上是递减，?f(x)max?f(1)?ln3?2?0. 由x?N*，则f(n)?f(1)?0，
故ln(1?bn)?bn. ????????????????????????????14分