Question

数列{an},a1=1,4a[n]*a[n+1]=(a[n]+a[n+1]-1)²,an>an-1,求an

Answer 1

a1=1>0 an>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.