已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}的通项公式谢谢

Question

已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}的通项公式
谢谢

Answer 1

Sn=1/8(an+2)^2
S(n-1)=1/8[a(n-1)+2]^2
2式相减整理得到
[an + a(n-1)]*[(1/4)(an - an-1) - 1] = 0
又an＞0
所以1/4(an-an-1)-1=0
an-an-1=4
所以an是个公差为4的等差数列
a1=1/8(a1+2)^2
8a1=a1^2+4a1+4
(a1-2)^2=0
a1=2
所以an=2+(n-1)*4=4n-2