已知数列{an}(n€N*)是首项为1的等差数列,其公差d>0,且a3,a7+2,3a9成等比数列

An=1+(n-1)d
A3=1+2d
A7+2=1+6d+2=3+6d
3A9=3(1+8d)
(A7+2)^2=A3×3A9
(3+6d)^2=(1+2d)×3(1+8d)
解方程得d=1或d=-1/2
因为d>0，所以舍去d=-1/2
则An=n

Sn=n(n+1)/2
f(n)=[n(n+1)/2]/[(n+18)*(n+1)(n+2)/2]
=n/(n+18)(n+2)
=n(n^2+20n+36)
=1/(n+20+36/n) (因为n+36/n>=2*√(n*36/n)=12
<=1/(20+12)
=1/32
则f(n)=Sn/(n+18)Sn+1得最大值=1/32
仅当切当n=36/n时，n=6等号成立

a3,a7+2,3a9成等比数列,有：
(1+6d+2)^2=(1+2d)(3(1+8d))
d=1,舍弃d=-1/2

1.通向公式an=n

2.楼主把这个式子写的明确一些好么？