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

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
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

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

构造4Sn-1=(an-1+1)^2;然后用4sn=（an+1）^2 -4Sn-1=(an-1+1)^2;整理得到。4an=an^2-(an-1)^2+2an-2(an-1).再整理得到，an^2-(an-1)^2=2(an-an-1)得到，an-(an-1)=2.即为等差，首项a1为1。公差为2。由于电脑打字不方便，有问题找我，QQ670637853