an是等差数列

an=a(n-1)+d
1/an*a(n-1)=1/d*{1/a(n-1)-1/[a(n-1)+d]}=1/d*(1/a(n-1)-1/an)

1/a1a2+1/a2a3+1/a3a4+...+1/an*a(n-1)
=1/d*{1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+1/a5-1/a5+...+1/a(n-1)-1/an}
=1/d*[1/a1-1/an]
=1/d*[(an-a1)/a1an]

题目有点错误:应为1/a1a2+1/a2a3+1/a3a4+...+1/an*a(n+1)=n/a1*a(n+1)
则:1/a1a2+1/a2a3+1/a3a4+...+1/an*a(n+1)=1/d[1/a1-1/a2+1/a2-1/a3+……-1/a（n+1）]=1/d[1/a1-1/a（n+1）]=1/d×[a（n+1）-a1]/a1a（n+1）=1/d×（a1+nd-a1）/a1a（n+1）=n/a1a（n+1）

1/an*a(n-1)=1/d * ( 1/an - 1/a(n+1) )
1/a1a2+1/a2a3+1/a3a4+...+1/an*a(n-1)=1/d * ( 1/a1 - 1/a(n+1) )
=n/a1*a(n+1)