Ta có :
\(\dfrac{3}{2}\) = \(\dfrac{3}{1.2}\) = 3 x \(\dfrac{1}{1.2}\) = 3 x ( 1 - \(\dfrac{1}{2}\) ) : 2 ;
\(\dfrac{3}{8}\) = \(\dfrac{3}{2.4}\) = \(\)3 x \(\dfrac{1}{2.4}\) = 3 x ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ) : 2 ;
\(\dfrac{3}{24}\) = \(\dfrac{3}{4.6}\) = 3 x \(\dfrac{1}{4.6}\) = 3. ( \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) ) : 2 ;
\(\dfrac{3}{9800}\) = \(\dfrac{3}{98.100}\) = 3 x \(\dfrac{1}{98.100}\) = 3 x ( \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) ) : 2 ;
\(\dfrac{3}{10200}\) = \(\dfrac{3}{100.102}\) = 3 x \(\dfrac{1}{100.102}\)= 3 x ( \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) )
Vậy : \(\dfrac{3}{2}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{24}\) + -- + \(\dfrac{3}{9800}\) + \(\dfrac{3}{10200}\)
= \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.4}\) + \(\dfrac{3}{4.6}\) +--+ \(\dfrac{3}{98.100}\) + \(\dfrac{3}{100.102}\)
= 3 x ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + --.+ \(\dfrac{1}{98.100}\) + \(\dfrac{1}{100.102}\) )
= 3 x ( \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) +=-.+ \(\dfrac{2}{98.100}\) + \(\dfrac{2}{100.102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) +-...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) + \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) ) : 2
= 3 x ( 1 - \(\dfrac{1}{102}\) ) : 2 = 3 x \(\dfrac{101}{102}\) : 2
= \(\dfrac{101}{68}\)
