(Ⅰ)由题意得an+1=[(an+1-an)+(an-an-1)+…+(a2-a1)]+a1=3(22n-1+22n-3+…+2)+2=22(n+1)-1.由此可知数列{an}的通项公式为an=22n-1.
(Ⅱ)由bn=nan=n•22n-1知Sn=1•2+2•23+3•25++n•22n-1,由此入手可知答案.
【解析】
(Ⅰ)由已知,当n≥1时,an+1=[(an+1-an)+(an-an-1)+…+(a2-a1)]+a1
=3(22n-1+22n-3+…+2)+2=22(n+1)-1.
而a1=2,
所以数列{an}的通项公式为an=22n-1.
(Ⅱ)由bn=nan=n•22n-1知Sn=1•2+2•23+3•25+…+n•22n-1①
从而22Sn=1•23+2•25+…+n•22n+1②
①-②得(1-22)•Sn=2+23+25+…+22n-1-n•22n+1.
即.