(1) (an+2)/2=根號(hào)下2Sn
所以8Sn=(an+2)^2
n=1,S1=a1.8a1=(a1+2)^2,得a1=2
n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
n=3,8S3=(a1+2)^2,8(a1+a2+a3)=(a3+2)^2,得a3=10
(2) 8Sn=(an+2)^2
當(dāng)n≥2時(shí),8S(n-1)=[a(n-1)+2]^2
兩式相減得8an=(an+2)^2-[a(n-1)+2]^2
(an)^2+4an+4-[a(n-1)]^2-4a(n-1)-4=8an
(an)^2-[a(n-1)]^2-4an-4a(n-1)=0
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-4]=0
∵ {an}是正數(shù)組成的數(shù)列,∴an>0,a(n-1)>0
∴ an-a(n-1)=4
∴ {an}是等差數(shù)列,首項(xiàng)為a1,公差為4.
∴ an=a1+(n-1)d=2+4(n-1)=4n-2