设公差为d,则d≠0.
a3、a6、a12成等比数列
a6²=a3×a12
(a2+4d)²=(a2+d)(a2+10d)
a2=2代入,整理,得
d²-d=0
d(d-1)=0
d=0(舍去)或d=1
a1=a2-d=2-1=1
an=a1+(n-1)d=1+(n-1)=n
1/[ana(n+2)]=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
1/(a1a3)+1/(a2a4)+...+1/[ana(n+2)]
=(1/2)[1/1-1/3+1/2-1/4+...+1/n -1/(n+2)]
=(1/2)[(1/1+1/2+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/2)[1/1+1/2-1/(n+1)-1/(n+2)]
=3/4 -1/[2(n+1)] -1/[2(n+2)]