Question

已知数列{an}满足1/a1a2+1/a2a3+...+1/a(n-1)an=n-1/a1an,求证:{an}是等差数列

急！马上就要！

Answer 1

证明：左边＝1/(a1a2)+1/(a2a3)+...+1/(an-1*an)
＝1/d(1/a1-1/a2)+1/d(1/a2-1/a3)+...+1/d(1/an-1-1/an)
=1/d[(1/a2-1/a1)+(1/a3-1/a2)+...+(1/an-1-1/an)]
=1/d[1/a1+(1/a2-1/a2)+(1/a3-1/a3)+...+(1/an-1 - 1/an-1)-1/an]
=1/d(1/a1-1/an)
=1/d *[(n-1)d/a1an]
=(n-1)/(a1an)＝右边

追问

。。。。求证：{an}为等差数列。不是求证前面一条式子。 已知数列{an}满足1/a1a2+1/a2a3+...+1/a(n-1)an=n-1/a1an(n≥3）,求证:{an}是等差数列。