(Ⅰ)解:∵数列{an}满足: ,2an+1=anan+1+1 ∴n=1时,2a2=a1a2+1,∴![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021548-45209.png) n=2时,2a3=a2a3+1,∴![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021548-18380.png) n=3时,2a4=a3a4+1,∴ ; (Ⅱ)猜想数列{an}的通项公式 , 证明:①当n=1,2,3,4时,由(Ⅰ)知结论成立; ②假设n=k时,结论成立,即 ,则n=k+1时,∴2an+1=anan+1+1 ∴ =![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021549-13091.png) 即n=k+1时,结论成立 由①②可知 ; (Ⅲ)解:由anbn=1﹣an,可得![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021550-32124.png) ∴S1+S2+…+Sn﹣1=(n﹣1)+![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021550-44218.png) =n+![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015021550-10251.png) +…+ ﹣1×(n﹣1) =n(1+ +…+ ﹣1) =n(Sn﹣1) |