数列an为等差数列a1=2,a1+a2+a3=12→
a1+a1+d+a1+2d=12,→3a1+3d=12,→3×2+3d=12,
∴公差d=2.
(1).an=a1+(n-1)d=2+(n-1)×2,→an=2n
Sn=n(a1+an)/2=n(2+2n)/2=n(n+1)
(2)1/S1+1/S2+1/S3+…+1/S10=
1/(1×2)+1/(2×3)+1/(3×4)+.+1/(10×11)=
[(1/1)-(1/2)]+[(1/2)-(1/3)]+.+[(1/10)-(1/11)]=
1-(1/11)=10/11