Giải câu 2 trang 32 sách toán VNEN lớp 6 tập 2

• Chứng minh: 

Ta có: $\frac{1}{n}\times\frac{1}{n\;+\;1} = \frac{1}{n(n+1)}$      (1)

Lại có: $\frac{1}{n} - \frac{1}{n\;+\;1} = \frac{1\times (n+1) - 1\times n}{n(n+1)} = \frac{1}{n(n+1)}$   (2)

Từ (1) và (2), suy ra: $\frac{1}{n}\times\frac{1}{n\;+\;1} = \frac{1}{n} - \frac{1}{n\;+\;1}$ (đpcm)

• Tính giá trị biểu thức:
• $A = \frac{1}{2}\times\frac{1}{3}\;+\; \frac{1}{3}\times\frac{1}{4}\;+\; \frac{1}{4}\times\frac{1}{5}\;+\; \frac{1}{5}\times\frac{1}{6}\;+\; \frac{1}{6}\times\frac{1}{7}\;+\; \frac{2}{7}\times\frac{1}{8}\;+\; \frac{1}{8}\times\frac{1}{9}$

$A = \frac{1}{2}-\frac{1}{3}\;+\; \frac{1}{3}-\frac{1}{4}\;+\; \frac{1}{4}-\frac{1}{5}\;+\; \frac{1}{5}-\frac{1}{6}\;+\; \frac{1}{6}-\frac{1}{7}\;+\; \frac{2}{7}-\frac{1}{8}\;+\; \frac{1}{8}-\frac{1}{9}$

$A = \frac{1}{2} - \frac{1}{9} = \frac{7}{18}$

• $B = \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110}$.

$B = \frac{1}{4}\times\frac{1}{5} + \frac{1}{5}\times \frac{1}{6} + \frac{1}{6}\times\frac{1}{7} + \frac{1}{7}\times\frac{1}{8} + \frac{1}{8}\times\frac{1}{9} + \frac{1}{9}\times\frac{1}{10} + \frac{1}{10}\times\frac{1}{11}$.

$B = \frac{1}{4}-\frac{1}{5} + \frac{1}{5}- \frac{1}{6} + \frac{1}{6}-\frac{1}{7} + \frac{1}{7}-\frac{1}{8} + \frac{1}{8}-\frac{1}{9} + \frac{1}{9}-\frac{1}{10} + \frac{1}{10}-\frac{1}{11}$.

$B = \frac{1}{4} - \frac{1}{11} = \frac{7}{44}$.