(an+2)/2=√(2Sn),
兩邊平方得：(an+2)²=8Sn,
n=1時(shí),(a1+2)²=8a1,a1=2.
(an+2)²=8Sn,(a(n-1)+2)²=8S (n-1). (n≥2)
兩式相減得：a²n+4 an- a²(n-1)-4 a(n-1)=8 an,
a²n- a²(n-1)- 4 an-4 a(n-1)=0
(an+ a(n-1))( an- a(n-1))-4(an+ a(n-1))=0,
an- a(n-1)=4,
所以數(shù)列{an}是首項(xiàng)為2,公差為4的等差數(shù)列.
an=2+4(n-1)=4n-2.