cn=3/[bnb(n+1)]=3/[(3n-2)(3(n+1)-2)]=3/[(3n-2)(3n+1)]=3×(1/3)×[1/(3n-2)-1/(3n+1)]=1/(3n-2)-1/[3(n+1)-2]Tn=c1+c2+...+cn=1/(3×1-2)-1/(3×2-2)+1/(3×2-2)-1/(3×3-2)+...+1/(3n-2)-1/[3(n+1)-2]=1/(3×1-2...3×(1/3)×[1/(3n-2)-1/(3n+1)]解释一下1/(3n-2) -1/[3(n+1)-2]=1/(3n-2)-1/(3n+1)=[(3n+1)-(3n-2)]/[(3n-2)(3n+1)]=3/[(3n-2)(3n+1)]1/[(3n-2)(3n+1)]=(1/3)[1/(3n-2) - 1/[3(n+1)-2] ]明白了吧。这是最基本的变形，会经常用到。 一般的：形如：1/[(an+b)(an+b+k)]=(1/k)[1/(an+b) - 1/(an +b+k)]