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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)
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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)