Ta có:
$B>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}$
Mà $\dfrac{1}{5}-\dfrac{1}{101}<\dfrac{1}5\Rightarrow B>\dfrac{1}5$
Ta có:
$B>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}$
Mà $\dfrac{1}{5}-\dfrac{1}{101}<\dfrac{1}5\Rightarrow B>\dfrac{1}5$
phải là $\dfrac{1}{6}<B$ chứ nhỉ? ^^
Ta có:
$B>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}-\dfrac{1}{30}=\dfrac{1}{6}$
phải là $\dfrac{1}{6}<B$ chứ nhỉ? ^^
Ta có:
$B>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{5}-\dfrac{1}{30}=\dfrac{1}{6}$