Giải thích các bước giải:
$\dfrac{1}{n(n-1)}=\dfrac{n-(n-1)}{n(n-1)}=\dfrac{1}{n-1}-\dfrac{1}{n}$
Ta có:
$\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{5.6}<\dfrac{1}{5^2}<\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}$
$\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{6.7}<\dfrac{1}{6^2}<\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}$
...
$\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{100.101}<\dfrac{1}{100^2}<\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}$
Cộng vế với vế ta được
$\dfrac{1}{5}-\dfrac{1}{101}<\dfrac{1}{5^2}+\dfrac{1}{6^2}+..+\dfrac{1}{100^2}<\dfrac{1}{4}-\dfrac{1}{100}$
Mà $\dfrac{1}{5}-\dfrac{1}{101}> \dfrac{1}{5}-\dfrac{1}{30}=\dfrac{1}{6}$
$\dfrac{1}{4}-\dfrac{1}{100}<\dfrac{1}{4}$
$\rightarrow \dfrac{1}{6}<\dfrac{1}{5^2}+\dfrac{1}{6^2}+..+\dfrac{1}{100^2}<\dfrac{1}{4}$
$\rightarrow đcpm$
