数列an各项都是正数,前n项为sn,且an和sn满足4sn=（an+1）^2 (n为正整数）,求证an是等差数列,并求a

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数列an各项都是正数,前n项为sn,且an和sn满足4sn=（an+1）^2 (n为正整数）,求证an是等差数列,并求an

a1=s1=(a1+1)^2/4得a1=1
n大于1时,an=sn-s（n-1）=(an+1)^2/4-(a（n-1）+1)^2/4
(an-1)^2=(a（n-1）+1)^2
若an-1=a（n-1）+1得an-a（n-1）=2
若-an+1=a（n-1）+1得an+a（n-1）=0（舍）
所以an是等差数列
an=1+2（n-1）=2n-1

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