Question

在数列{an}中,a1=1,an+1=an+c(c为常数,n∈N*),且a1,a2,a5成公比不为1的等比数列.

Answer 1

a(n+1)=an+c

a(n+1)-an=c,为定值,又a1=1,数列{an}是以1为首项,c为公差的等差数列

a1、a2、a5成等比数列,则

a2²=a1·a5

(a1+c)²=a1(a1+4c)

c²-2a1c=0

a1=1代入,得c(c-2)=0

若c=0,则a1=a2=a5,公比为1,与已知矛盾,舍去

c=2

an=a1+c(n-1)=1+2(n-1)=2n-1

bn=1/[ana(n+1)]=1/[(2n-1)(2(n+1)-1)]=(1/2)[1/(2n-1)-1/(2(n+1)-1)]

Sn=b1+b2+...+bn

=(1/2)[1/(2×1-1)-1/(2×2-1)+1/(2×2-1)-1/(2×3-1)+...+1/(2n-1)-1/(2(n+1)-1)]

=(1/2)[1- 1/(2n+1)]

=n/(2n+1)