已知数列an的首项为1的等差数列,且an+1>an,若a3,a7+,3a9成等比数列

an=1+(n-1)da(n+1)>an即1+nd>1+(n-1)d所以d>0
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=a1+(n-1)=1+(n-1)*1=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)