求解过程如图,满意请采纳。
![](https://iknow-pic.cdn.bcebos.com/838ba61ea8d3fd1fa4b72946204e251f95ca5f14?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(I)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3
两式相减得an+12-an2+2(an+1-an)=4an+1,
即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an),
∵an>0,∴an+1-an=2,
∵a12+2a1=4a1+3,
∴a1=-1(舍)或a1=3,
则{an}是首项为3,公差d=2的等差数列,
∴{an}的通项公式an=3+2(n-1)=2n+1:
(Ⅱ)∵an=2n+1,
∴bn===(-),
∴数列{bn}的前n项和Tn=(-+…+-)=(-)=.