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考察一般项第k项：
1/[aka(k+1)]=1/[(3k-2)(3(k+1)-2)]=1/[(3k-2)(3k+1)]=(1/3)[1/(3k-2)-1/(3k+1)]
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/3)[1/1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)]
=(1/3)[1-1/(3n+1)]
=n/(3n+1)
1/[aka(k+1)]=1/[(3k-2)(3(k+1)-2)]=1/[(3k-2)(3k+1)]=(1/3)[1/(3k-2)-1/(3k+1)]
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/3)[1/1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)]
=(1/3)[1-1/(3n+1)]
=n/(3n+1)
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Sn=(1-1/4+1/4-1/7+1/7-1/10+......+1/(3n-2)-1/(3n+1))/3
=(1-1/(3n+1))/3=n/(3n+1)
=(1-1/(3n+1))/3=n/(3n+1)
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