填空题设Sn是正项数列{an}的前n项和,且an和Sn满足:,则Sn=________.
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n2解析分析:利用数列递推式,再写一式,两式相减,求出数列的通项,即可得到结论.解答:∵4Sn=(an+1)2,∴n≥2时,4Sn-1=(an-1+1)2,作差,得4(Sn-Sn-1)=(an+1)2-(an-1+1)2,∴4an=(an+an-1+2)(an-an-1),整理,得(an+an-1)(an-an-1-2)=0.∵{an}正数数列,∴an-an-1=2,∵4S1=(a1+1)2,∴a1=1,∴an=2n-1∴4Sn=(2n-1+1)2,∴Sn=n2,故