题目

{an}是等差数列,an≠0,求1/a1a2+1/a2a3+.+1/a(n-1)an
{an}是等差数列,an≠0,求1/a1a2 + 1/a2a3 + .+ 1/a(n-1)an
运用列项相消法做

提问时间：2020-11-16

答案

设a1=a
则1/a1a2 + 1/a2a3 + .+ 1/a(n-1)an
=1/a(a+d)+1/(a+d)(a+2d)+……+1/[a+(n-2)d][a+(n-1)d]
={d/a(a+d)+d/(a+d)(a+2d)+……+d/[a+(n-2)d][a+(n-1)d]}/d
={1/a-1/(a+d)+1/(a+d)-1/(a+2d)+……+1/[a+(n-2)d]-1/[a+(n-1)d]}/d
={1/a-1/[a+(n-1)d]}/d
=[1/a-1/(a+nd-d)]/d
=(a+nd-d-a)/d(a+nd-d)
=(nd-d)/d(a+nd-d)
=(n-1)/(a+nd-d)