若数列{an}是等差数列,an≠0,则【1/a1a2】+【1/a2a3】+`````+【1/a(n-1)an】=?

题目

若数列{an}是等差数列,an≠0,则【1/a1a2】+【1/a2a3】+`````+【1/a(n-1)an】=?

答案

设公差为d,则【1/a1a2】+【1/a2a3】+`````+【1/a(n-1)an】
＝(1/d)*[(1/a1－1/a2)+(1/a2-1/a3)+(1/a3-1/a4)+...+(1/a(n-1)-1/an)]
=1/d*(1/a1-1/an)
=1/d*(an-a1)/(an*a1)
=1/d*(n-1)d/(an*a1)
=n-1/(an*a1)