Question

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

Answer 1

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)