正项数列{an}的前n项和为Sn,且Sn=(an+2)^2/8,求an的通项公式?

题目

正项数列{an}的前n项和为Sn,且Sn=(an+2)^2/8,求an的通项公式?

答案

Sn=(an+2)^2/8当n=1时,a1=S1=(a1+2)^2/8∴(a1)^2+4a1+4=8a1(a1)^2-4a1+4=0那么a1=2又S(n+1)=[a(n+1)+2]^2/8所以a(n+1)=S(n+1)-Sn=(an+2)^2/8-[a(n+1)+2]^2/8∴8a(n+1)=(an+2)^2-[a(n+1)+2]^2[a(n+1)]^2-4a(n+1)+4=(a...