问题标题：
设{an}是等差数列,且首项a1>0,公差d>0求证:1/a1a2+1/a2a3+…+1/anan+1=n/a1(a1+nd)
问题描述：
设{an}是等差数列,且首项a1>0,公差d>0求证:1/a1a2+1/a2a3+…+1/anan+1=n/a1(a1+nd)
姜重然回答：
1/a1a2+1/a2a3+…+1/anan+1
=[(a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1]/d
=[1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1)]/d
=[1/a1-1/a(n+1)]/d
=(a(n+1)-a1)/a1a(n+1)d
=nd/a1a(n+1)d
=n/a1(a1+nd)
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