(1)∵an+1=2an+1,
∴an+1+1=2(an+1).
又a1=1,
故数列{an+1}是首项为2,公比为2的等比数列.
∴an+1=2n,
an=2n−1;
(2)∵[2n
an+1=
2n
2n=
n
2n−1,
∴Sn=1+
2/2+
3
22+…+
n
2n−1],
[1/2Sn=
1
2+
2
22+
3
23+…+
n−1
2n−1+
n
2n],
以上两式相减,得
[1/2Sn=1+
1
2+
1
22+…+
1
2n−1−
n
2n]=
1−(
1
2)n
1−
1
2−
n
2n=2−
n+2
2n.
∴Sn=4−
n+2
2n−1;
(3)证明:∵
1
an+1=
1