求此数列题详细解法已知数列an Sn=1/8(an+2)^2 求证an是等差数列

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求此数列题详细解法
已知数列an Sn=1/8(an+2)^2 求证an是等差数列

Sn=1/8*(an+2)^2
Sn+1=1/8*(an+1+2)^2
Sn+1-Sn=1/8*[(an+1^2+4an+1)-(an^2+4an)]
8an+1=(an+1^2+4an+1)-(an^2+4an)
(an+1^2-4an+1+4)-(an^2+4an+4)=0
(an+1-2)^2-(an+2)^2=0
(an+1-2)^2=(an+2)^2
an+1-2=an+2或an+1-2=-an-2
an+1-an=4或an+1+an=0

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