11. 新考向 阅读理解题 阅读材料,并回答问题:
计算$(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$时,若把$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$与$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$分别看成一个整体,再利用分配律进行计算,可以大大降低算式的复杂度.过程如下:
解:设$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$为$A$,$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$为$B$,则原式$=(1+A)× B-(1+B)× A=B+A× B - A - A× B = B - A = \frac{1}{5}$.
请用上面的方法计算:
(1) $(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})×(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})$.
(2) $(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n})×(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n+1})-(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n+1})×(\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n})$.

11. (1) 设$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$为$A$,$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$为$B$,则原式$=(1+A)×B-(1+B)×A=B+A×B-A-A×B=B-A=\frac{1}{7}.$
(2) 设$\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}$为$A$,$\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n+1}$为$B$,则原式$=(1+A)×B-(1+B)×A=B+A×B-A-A×B=B-A=\frac{1}{n+1}.$