已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为
问题描述：
已知等差数列公差为d,1/a1a2+1/a2a3+…+1/anan+1可化简为
已知等差数列公差为d,且a1≠0,d≠0,则1/a1a2+1/a2a3+…+1/anan+1可化简为 n/a1(a1+nd)求详细步骤
答
因为1/anan+1=1/an*（an+d）=1/d[1/an-1/（an+d）]=1/d[1/an-1/an+1]
所以1/a1a2+1/a2a3+…+1/anan+1
=1/d[1/a1-1/a2+1/a3-1/a4.1/an-1/an+1]
=1/d（1/a1-1/an+1）
=1/d*（an+1-a1）/a1an+1
=n/a1(a1+nd)