Bài 17:
a, $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + ... + $\frac{1}{1999.2000}$
= $\frac{2-1}{1.2}$ + $\frac{3-2}{2.3}$ + ... + $\frac{2000-1999}{1999.2000}$
= 1 - $\frac{1}{2}$ +$\frac{1}{2}$ - $\frac{1}{3}$ +... + $\frac{1}{1999}$ - $\frac{1}{2000}$
= 1 - $\frac{1}{2000}$
= $\frac{1999}{2000}$
b, $\frac{1}{1.4}$ + $\frac{1}{4.7}$ + ... + $\frac{1}{100.103}$
= $\frac{1}{3}$.($\frac{3}{1.4}$ + $\frac{3}{4.7}$ + ... + $\frac{3}{100.103}$
= $\frac{1}{3}$.($\frac{4-1}{1.4}$ + $\frac{7-4}{4.7}$ + ... + $\frac{103-100}{100.103}$
= $\frac{1}{3}$ . (1 - $\frac{1}{103}$)
= $\frac{1}{3}$ . $\frac{102}{103}$
= $\frac{34}{103}$
c, -$\frac{1}{2000.1999}$ - $\frac{1}{1999.1998}$ - ... - $\frac{1}{2.1}$
= -($\frac{1}{1.2}$ + ... + $\frac{1}{1998.1999}$ + $\frac{1}{1999.2000}$)
= -(1 - $\frac{1}{2000}$)
= -$\frac{1999}{2000}$
d, $\frac{-1}{3}$ + $\frac{-1}{15}$ + ... + $\frac{-1}{9999}$
= -($\frac{1}{3}$ + $\frac{1}{15}$ + ... + $\frac{1}{9999}$)
= -($\frac{1}{1.3}$ + $\frac{1}{3.5}$ + ... + $\frac{1}{99.101}$)
= -$\frac{1}{2}$ . ($\frac{2}{1.3}$ + $\frac{2}{3.5}$ + ... + $\frac{2}{99.101}$)
= -$\frac{1}{2}$ . (1 - $\frac{1}{101}$)
= -$\frac{1}{2}$ . $\frac{100}{101}$
= -$\frac{50}{101}$
e, $\frac{8}{9}$ - $\frac{1}{72}$ - $\frac{1}{56}$ - $\frac{1}{42}$ - ... - $\frac{1}{6}$ - $\frac{1}{2}$
= $\frac{8}{9}$ - ($\frac{1}{1.2}$ + $\frac{1}{2.3}$ + ... + $\frac{1}{8.9}$)
= $\frac{8}{9}$ - $\frac{8}{9}$
= 0
f, 1 - $\frac{1}{2.5}$ - $\frac{1}{5.8}$ - ... - $\frac{1}{92.95}$
= $\frac{1}{3}$ - ($\frac{3}{2.5}$ + $\frac{3}{5.8}$ + ... + $\frac{3}{92.95}$)
= $\frac{1}{3}$ - ($\frac{1}{2}$ - $\frac{1}{95}$)
= $\frac{1}{3}$ - $\frac{93}{190}$
= -$\frac{89}{570}$
