已知数列{an}的前n项和为Sn，且Sn=2an-2，求通项公式。（急速）
\u5df2\u77e5\u6570\u5217{an}\u7684\u524dN\u9879\u548c\u4e3aSn \u4e14an+1=Sn-n+3,a1=2,.\u6c42an\u7684\u901a\u9879\u516c\u5f0f\u4f60\u597d
\u6211\u8ba4\u4e3a\u95ee\u9898\u51fa\u5728n\u7684\u53d6\u503c\u8303\u56f4\u4e0a
an+1=Sn-n+3,\u6b64\u65f6n\u7684\u53d6\u503c\u8303\u56f4\u662fn\u22651
an=Sn-1-n+4\u6b64\u65f6n\u7684\u53d6\u503c\u8303\u56f4\u662fn\u22652
\u6240\u4ee5\u4e24\u8005\u5fc5\u987b\u5148\u7edf\u4e00n\u7684\u8303\u56f4\u624d\u80fd\u591f\u76f8\u51cf
\u6240\u4ee5\u5e94\u8be5\u4e3a2an-1=an+1(n\u22652\uff09
\u6545\u4e0d\u80fd\u5c06n=1\u4ee3\u5165\u4e0a\u5f0f\u3002
2an-1=an+1(n\u22652\uff09
\u5373(an+1-1)-2(an-1)=0
\u4ee4an-1=bn
\u5219bn+1-2bn=0
(bn+1)/bn=2
\u6545bn\u662f\u7b49\u6bd4\u6570\u5217
b1=a1-1=1
\u6240\u4ee5bn=2^(n-1)
\u6240\u4ee5an=bn+1=2^(n-1)+1(n\u22652\uff09
\u8fd9\u65f6\u624d\u80fd\u5c06n=1\u4ee3\u5165\u68c0\u9a8c
a1=1+1=2\uff0c\u7b26\u5408\u4e0a\u5f0f
\u6240\u4ee5an=2^(n-1)+1
\u89e3\uff1a
a1=S1=1^2+1=2
Sn=n^2+1
Sn-1=(n-1)^2+1
an=n^2+1-(n-1)^2-1=2n-1
n=1\u65f6\uff0ca1=1\uff0c\u4e0ea1=2\u77db\u76fe\uff0cn=1\u65f6\uff0ca1=2
\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3a
an=2 (n=1)
=2n-1 (n>1)
那么
S(n+1)=2a(n+1)-2
两式相减
S(n+1)-Sn=2a(n+1)-2an
a(n+1)=2a(n+1)-2an
a(n+1)=2an
所以an是以2为公比的等比数列
当n=1时
s1=2a1-2=a1
所以a1=2
所以an=2^n
∵Sn=2an - 2
Sn-1=2an-1 - 2
两式相减,得：an = 2an - 2an-1
-an=-2an-1
∴an / an-1 = 2
∴数列{an}是公比为2的等比数列
S1=2a1-2
∵S1=a1
∴a1=2
an=2×2^(n-1) =2^n
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