(Ⅰ)证明:∵2=an+1, ∴, ∴a n+1=S n+1﹣Sn=﹣=, 即:2(a n+1+an)=(an+1+an)(an+1﹣an), ∴(an+1+an)(an+1﹣an﹣2)=0, ∵an>0,∴a n+1+an>0, ∴an+1﹣an﹣2=0, ∴a n+1﹣an=2, 当n=1时,S1=,即a1=, ∴,解得a1=1. ∴数列{an}是首项为a1=1,公差d=2的等差数列. (Ⅱ)解:由(Ⅰ)知:an=a1+(n﹣1)d=2n﹣1, ∵=, ∴Tn=b1+b2+…+bn=, ① =, ② ①﹣②得:==, ∴. (Ⅲ)证明:由(Ⅱ)得:=, ∴==, ∴c1+c2+c3+…+cn= =, 故,. |