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
![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](/uploads/image/z/18485901-45-1.jpg?t=%E6%95%B0%E5%88%97an%E5%90%84%E9%A1%B9%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%2C%E5%89%8Dn%E9%A1%B9%E4%B8%BAsn%2C%E4%B8%94an%E5%92%8Csn%E6%BB%A1%E8%B6%B34sn%3D%EF%BC%88an%2B1%EF%BC%89%5E2+%28n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%2C%E6%B1%82%E8%AF%81an%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82a)
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
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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