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) 1个回答 分类:数学 2014-12-16 问题解答: 我来补答 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) 展开全文阅读