数列{an},a1=1,4a[n]*a[n+1]=(a[n]+a[n+1]-1)²,an>an-1,求an
问题描述：
数列{an},a1=1,4a[n]*a[n+1]=(a[n]+a[n+1]-1)²,an>an-1,求an
答
a1=1>0an>a(n-1) ,数列为递增数列,各项均为正.
√a(n+1)>√an
4ana(n+1)=[an+a(n+1) -1]²
2√[ana(n+1)=an+a(n+1)-1
a(n+1)-2√[ana(n+1)]+an=1
[√a(n+1)-√an]²=1
√a(n+1)-√an=1,为定值.
√a1=√1=1
数列{√an}是以1为首项,1为公差的等差数列.
√an=1+n-1=n
an=n²
n=1时,a1=1²=1,同样满足.
数列{an}的通项公式为an=n².