数列{an}满足a1=1⼀3,当n属于正整数且n>1时,有a(n-1)·an=a(n-1)-an

2025-01-05 16:40:04
推荐回答(1个)
回答1:

1.
a(n-1)an=a(n-1)-an
等式两边同除以ana(n-1)
1/an -1/a(n-1)=1,为定值。
1/a1=1/(1/3)=3
数列{1/an}是以3为首项,1为公差的等差数列。
bn=1/an,数列{bn}是以3为首项,1为公差的等差数列。
bn=3+(n-1)=n+2
n=1时,b1=1/(a1)=1/(1/3)=3 b1=1+2=3,同样满足。
数列{bn}的通项公式为bn=n+2。
2.
an=1/bn=1/(n+2)
an/n=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
Tn=a1/1+a2/2+...+an/n
=(1/2)[1/1-1/3+1/2-1/4+...+1/n -1/(n+2)]
=(1/2)[(1+1/2+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2]
=3/4 -1/[2(n+1)] -1/[2(n+2)]

Tn-[3/4 -1/(n+2)]
=3/4 -1/[2(n+1)]-1/[2(n+2)] -3/4 +2/[2(n+1)]
=1/[2(n+1)-1/[2(n+2)]
=(1/2)[1/(n+1)-1/(n+2)]
=1/[2(n+1)(n+2)]>0
Tn>3/4 -1/(n+2)