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∵Sn=n2-2n+3,a1=2,∴an=Sn-Sn-1=n2-2n+3-[(n-1)2-2(n-1)+3]=2n-3(n>1),∵當n=1時,a1=-1≠2,∴an=2,n=12n−3,n>1.,故答案為an=2,n=12n−3,n>1.
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