求数列通项:已知正数数列{an}其前n项和为Sn满足Sn=1/4[(an+1)^2],求an

Question

求数列通项:已知正数数列{an}其前n项和为Sn满足Sn=1/4[(an+1)^2],求an

Answer 1

an=Sn-S(n-1)
=1/4(an+1)²-1/4[a(n-1)+1]²
an=1/4(an²+2an+1)-1/4[a(n-1)+1]²
1/4(an²+2an+1)-an=1/4[a(n-1)+1]²
1/4(an²-2an+1)=1/4[a(n-1)+1]²
1/4(an-1)²=1/4[a(n-1)+1]²
所以an-1=a(n-1)+1
或an-1=-[a(n-1)+1]
若an-1=-[a(n-1)+1]
an=-a(n-1)
an是正数,所以不成立
所以an-1=a(n-1)+1
an-a(n-1)=2
所以是等差数列
d=2
a1=S1
所以a1=1/4(a1+1)²
(a1-1)²=0
a1=1
所以an=2n-1