Question

在数列{an}中,an≠0,a1=1/3,并且对任意n∈N*,n≥2都有An·An-1=An-1-An令bn=1/an(n∈N*)

(1)求数列{bn}的通项公式
(2)求数列{an/n}的前n项的和Tn

Answer 1

a(n+1)a(n)=a(n)-a(n+1),
a(1)=1/3>0, a(n)>0.
1=1/a(n+1)-1/a(n)=b(n+1)-b(n),
{b(n)}是首项为1/a(1)=3,公差为1的等差数列.
b(n)=3+(n-1)=n+2
a(n)=1/b(n)=1/(n+2)
a(n)/n = 1/[n(n+2)] = (1/2){1/[n(n+1)] + 1/[(n+1)(n+2)]}=(1/2)[1/n - 1/(n+1)] + (1/2)[1/(n+1)-1/(n+2)]
t(n)=a(1)/1 + a(2)/2 + ... + a(n-1)/(n-1) + a(n)/n
=(1/2)[1-1/2+1/2-1/3+...+1/(n-1)-1/n+1/n-1/(n+1)] + (1/2)[1/2-1/3+1/3-1/4+...+1/n-1/(n+1)+1/(n+1)-1/(n+2)]
=(1/2)[1-1/(n+1)] + (1/2)[1/2-1/(n+2)]
=3/4 - (1/2)[1/(n+1)+1/(n+2)]