题目

数列an满足a1=1/4,a2=1/5,且a1a2+a2a3+...+anan+1
数列{an}满足a1=1/4,a2=1/5,且a1a2+a2a3+…+anan+1=na1an+1对于任何正整数n都成立,则1/a1+1/a2+...+1/a97的值为

提问时间：2020-10-31

答案

a1a2+a2a3+…+ana(n+1)=na1a(n+1)
a1a2+a2a3+…+a(n-1)ana=(n-1)a1an
两式相减得
ana(n+1)=na1a(n+1)-(n-1)a1an
等式两边同时除以ana(n+1)
1=na1/an-(n-1)a1/a(n+1)
1/a1=n/an-(n-1)/a(n+1)
n/an-(n-1)/a(n+1)=1/(1/4)
n/an-(n-1)/a(n+1)=4.1
同理得
(n-1)/a(n-1)-(n-2)/an=4.2
1式-2式得
2(n-1)/an-(n-1)/a(n+1)-(n-1)/a(n-1)=0
2(n-1)/an=(n-1)/a(n+1)+(n-1)/a(n-1)
2/an=1/a(n+1)+1/a(n-1)
所以1/an是等差数列．
d=1/a2-1/a1
=1/(1/5)-1/(1/4)
=5-4
=1
1/an=1/a1+(n-1)d
=1/(1/4)+n-1
=4+n-1
=n+3
1/a1+1/a2+...+1/a97
=4+5+.+100
=(4+100)*97/2
=5044