{an}为等差数列,an不等于0,d为公差,求证:1/(a1a2)+1/(a2a3)+...+1/(an-1*an)=(
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{an}为等差数列,an不等于0,d为公差,求证:1/(a1a2)+1/(a2a3)+...+1/(an-1*an)=(n-1)/(aian)
![{an}为等差数列,an不等于0,d为公差,求证:1/(a1a2)+1/(a2a3)+...+1/(an-1*an)=(](/uploads/image/z/10447886-38-6.jpg?t=%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Can%E4%B8%8D%E7%AD%89%E4%BA%8E0%2Cd%E4%B8%BA%E5%85%AC%E5%B7%AE%2C%E6%B1%82%E8%AF%81%3A1%2F%28a1a2%29%2B1%2F%28a2a3%29%2B...%2B1%2F%28an-1%2Aan%29%3D%28)
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
=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
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