数列an中,已知a1an>0,Sn是数列an的前n项和,2Sn=p(2an^2+an-1)

1.n=1时,2S1=2a1=p(2a1^2+a1-1),所以p=1
故 2Sn=2an^2+an-1
n=2时,2S2=2（a1+a2)=2a2^2+a2-1 ,故 a2=3/2
同理 a3=2,
2.2Sn=2an2+an-1
2Sn-1=2an-1^2+an-1 -1
上式减下式,有 2an=2(an^2-an-1^2)+(an-an-1)
整理：an+an-1=2(an^2-an-1^2)=2(an+an-1)(an-an-1)
因为：an+an-1=0时,不合题意；
即an-an-1=1/2=d
公差为1/2 ,首项为1,故an=1+(n-1)/2=(n+1)/2

因2a1^2-a1-1=0 an>0 所以a1=1又2S1=P(2a1^2 a1-1)所以P=1 2Sn=2an^2 an-1所以Sn S〈n-1〉=2an^2-1所以S〈n 1〉 Sn=2a〈n 1〉^2-1所以a〈n 1〉 an=2(a〈n 1〉 an)(a〈n 1〉-an)所以an=1 (n-1)/2

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