Question

已知数列{an}的前n项和为Sn,且Sn=2–(2/n+1)an 求证：数列{an/n}是等比数列,

Answer 1

a1=S1=2-[(2/1)+1]a1

整理,得4a1=2 a1=1/2

n≥2时,

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

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

Sn-Sn-1=an=2-[(2/n)+1]an-2+[2/(n-1) +1]a(n-1)

整理,得

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

2an/n=a(n-1)/(n-1)

(an/n)/[a(n-1)/(n-1)]=1/2,为定值.

a1/1=(1/2)/1=1/2

数列{an/n}是以1/2为首项,1/2为公比的等比数列.

an/n=(1/2)×(1/2)^(n-1)=1/2ⁿ