Câu 2 :
$\dfrac{3}{1.4} + \dfrac{3}{4.7} + ... + \dfrac{3}{97.100} < 1$
$= 1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + ... + \dfrac{1}{97} - \dfrac{1}{100} < 1$
$= 1 - \dfrac{1}{100} < 1$
$= \dfrac{99}{100} < 1$
$-> ĐPCM$
Câu 3 :
$\dfrac{3^2}{1.4} + \dfrac{3^2}{4.7} + ... + \dfrac{3^2}{97.100}$
$= 3 ( \dfrac{3}{1.4} + \dfrac{3}{4.7} + ... + \dfrac{3}{97.100} )$
$= 3 ( 1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + ... + \dfrac{1}{97} - \dfrac{1}{100} )$
$= 3 ( 1 - \dfrac{1}{100} )$
$= 3 . \dfrac{99}{100}$
$= \dfrac{297}{100}$
