Question

设正数列{an}的前n项和为Sn，且2*根号下Sn=an+1，试求数列{an}的通项公式。

求证{an}是等比数列

Answer 1

将 n=1，2，3 代入 2√Sn = an+1，得2√a1 = a1 +1 ，得 a1 =12√(a1+a2) = a2 +1 得 a2 = 32√(a1+a2+a3) = a3 +1 得 a3 =5 ……an = 2n-1 是等差数列啊

Answer 2

2√Sn=an+1则Sn=(an^2+2an+1)/4an=Sn-S(n-1)an=(an^2+2an+1)/4-(a(n-1)^2-2a(n-1)+1)/4an=(an^2-a(n-1)^2+2an-2a(n-1))/44an=an^2-a(n-1)^2+2an-2a(n-1)2an+2a(n-1)=(an+a(n-1))(an-a(n-1))an-a(n-1)=2a1=S1=1{an}为等差数列an=2n-1