已知等差数列an首项为a1,公差为d,an≠0，求数列{1/[a（n+1）*an]}前n项和,用d和a1表示
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an=a1+d(n-1)
1/[a(n+1)*an]=d[1/(a1+dn-d)-1/(a1+d*n)]
1/(a2*a1)+1/(a3*a2)+....+1/(a1+dn)*(a1+dn-d)
=d(1/a1-1/(a1+d)+1/(a1+d)-1/(a1+2d)+...+1/(a1+dn-2d)-1/(a1+dn-d)+1/(a1+dn+d)-1/(a1+dn)
=d/(1/a1-1/(a1+d*n)
1/[a(n+1)*an]=d[1/(a1+dn-d)-1/(a1+d*n)]
1/(a2*a1)+1/(a3*a2)+....+1/(a1+dn)*(a1+dn-d)
=d(1/a1-1/(a1+d)+1/(a1+d)-1/(a1+2d)+...+1/(a1+dn-2d)-1/(a1+dn-d)+1/(a1+dn+d)-1/(a1+dn)
=d/(1/a1-1/(a1+d*n)
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