{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)

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

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