已知数列{an}满足:1•a1+2•a2+3•a3+…n•an=n
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已知数列{an}满足:1•a1+2•a2+3•a3+…n•an=n
(1)求{an}的通项公式;
(2)若b
(1)求{an}的通项公式;
(2)若b
(1)∵数列{an}满足:1•a1+2•a2+3•a3+…n•an=n,
∴当n≥2时,nan=
n
i=1i•ai-
n−1
ii•ai=1,
∴an=
1
n,
当n=1时,a1=1成立,
∴an=
1
n.
(2)∵bn=n•2n,
∴Sn=1•21+2•22+3•23…+n•2n①
2Sn=1•22+2•23+3•24…+(n−1)•2n+n•2n+1②
由①-②得,−Sn=21+22+23+…+2n−n•2n+1
=
2(1−2n)
1−2−n•2n+1=(1−n)•2n+1−2,
∴Sn=(n−1)•2n+1+2.
∴当n≥2时,nan=
n
i=1i•ai-
n−1
ii•ai=1,
∴an=
1
n,
当n=1时,a1=1成立,
∴an=
1
n.
(2)∵bn=n•2n,
∴Sn=1•21+2•22+3•23…+n•2n①
2Sn=1•22+2•23+3•24…+(n−1)•2n+n•2n+1②
由①-②得,−Sn=21+22+23+…+2n−n•2n+1
=
2(1−2n)
1−2−n•2n+1=(1−n)•2n+1−2,
∴Sn=(n−1)•2n+1+2.
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