解:各项均为正数的数列{an}满足a1=1,且.
∴an+1·an(an+1+an)+(an+1+an)(an+1﹣an)=0
(an+1+an)(an+1·an+an+1﹣an)=0
∴an+1·an+an+1﹣an=0
∴+1=0;
∴=1.①
(Ⅰ)∵=1+=2
∴a2=;同理:a3=.
(Ⅱ)由①得是首项为1,公差为1的等差数列;
∴=1+(n﹣1)×1=n;
∴an=.
(Ⅲ)∵=2n+;
{n·2n}的和Sn=1·21+2·22+…+n·2n …①,
2·Sn=2·21+3·22+…+n·2n+1 …②,
∴①﹣②得﹣Sn=21+22+23+…+2n﹣n·2n+1
∴﹣Sn=﹣n×2n+1
∴Sn=(n﹣1)2n+1+2;
{}的和为:Tn=(1﹣)+()+…+()=1﹣=.
∴数列{bn}的前n项和为:Sn+Tn=(n﹣1)2n+1+2+.
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