a)Tính M=1/3 + 1/6 + 1/10 + 1/15 +……+ 1/2004.2005 b)Tính S= 1/1x2x3 + 1/2x3x4 + 1/3x4x5 + …..+ 1/98.99.100

Giải thích các bước giải:

$M = \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + … + \frac{1}{2004.2005}$

⇔$\frac{1}{2}M = \frac{1}{6} + \frac{1}{12} + … + \frac{1}{2004.2005}$

⇔$\frac{1}{2}M = \frac{1}{2.3} + \frac{1}{3.4} + … + \frac{1}{2004.2005}$

⇔$\frac{1}{2}M = \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} +  … + \frac{1}{2004} – \frac{1}{2005}$

⇔$\frac{1}{2}M = \frac{1}{2} – \frac{1}{2005}$

⇔$\frac{1}{2}M = \frac{2003}{4010}$

⇔$M = \frac{2003}{4010} : \frac{1}{2}$

⇔$M = \frac{2003}{2005}$