设数列满足a1=2,an+1-an=3•22n-1
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设数列满足a1=2,an+1-an=3•22n-1
(1)求数列{an}的通项公式;
(2)令bn=nan,求数列{bn}的前n项和Sn.
(1)求数列{an}的通项公式;
(2)令bn=nan,求数列{bn}的前n项和Sn.
(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Sn=
1
9[(3n−1)22n+1+2].
=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Sn=
1
9[(3n−1)22n+1+2].
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