(2012•道里区三模)已知数列{an}的前n项和为Sn,满足Sn=n2an−n2(n−1),且a1=12.
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(2012•道里区三模)已知数列{an}的前n项和为Sn,满足S
(1)∵Sn=n2an−n2(n−1),且a1=
1
2,
∴当n≥2时,有an=Sn-Sn-1,
∴Sn=n2(Sn−Sn−1)-n2(n-1),
即(n2-1)Sn=n2Sn−1+n2(n−1),
∵bn=
n+1
nSn,∴Sn=
n
n+1bn,
从而bn-bn-1=n.
(2)由(1)知
bn-b1=n+(n-1)+…+2=
n(n+1)
2−1,
b1=2S1=1,
∴bn=
n(n+1)
2,
∴Sn=
n
n+1•bn=
n
n+1•
n(n+1)
2=
n2
2,
∴a1=S1=
1
2,
an=Sn-Sn-1=
n2
2−
(n−1)2
2=
2n−1
2,
当n=1时,
2n−1
2=
1
2,
故an=
2n−1
2.
1
2,
∴当n≥2时,有an=Sn-Sn-1,
∴Sn=n2(Sn−Sn−1)-n2(n-1),
即(n2-1)Sn=n2Sn−1+n2(n−1),
∵bn=
n+1
nSn,∴Sn=
n
n+1bn,
从而bn-bn-1=n.
(2)由(1)知
bn-b1=n+(n-1)+…+2=
n(n+1)
2−1,
b1=2S1=1,
∴bn=
n(n+1)
2,
∴Sn=
n
n+1•bn=
n
n+1•
n(n+1)
2=
n2
2,
∴a1=S1=
1
2,
an=Sn-Sn-1=
n2
2−
(n−1)2
2=
2n−1
2,
当n=1时,
2n−1
2=
1
2,
故an=
2n−1
2.
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