Đáp án:
Giải thích các bước giải:
a) $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{1999.2000}$
=$\frac{2-1}{1.2}$ + $\frac{3-2}{2.3}$ + $\frac{4-3}{3.4}$ + ... + $\frac{2000-1999}{1999.2000}$
=$\frac{2}{1.2}$ - $\frac{1}{1.2}$ + $\frac{3}{2.3}$ - $\frac{2}{2.3}$ + ... + $\frac{2000}{1999.2000}$ - $\frac{1999}{1999.2000}$
=$\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... +
$\frac{1}{1999}$ - $\frac{1}{2000}$
= 1- $\frac{1}{2000}$
= $\frac{1999}{2000}$
b) $\frac{1}{1.4}$ + $\frac{1}{4.7}$ + $\frac{1}{7.10}$ + ... + $\frac{1}{100.103}$
= $\frac{4-1}{1.4}$ + $\frac{7-4}{4.7}$ + $\frac{10-7}{7.10}$ + ... + $\frac{103-100}{100.103}$
= $\frac{4}{1.4}$ - $\frac{1}{1.4}$ + $\frac{7}{4.7}$ - $\frac{4}{4.7}$ + ... + $\frac{103}{100.103}$ - $\frac{100}{100.103}$
= $\frac{1}{1}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{7}$ + ... + $\frac{1}{100}$ - $\frac{1}{100}$
= 1 - $\frac{1}{103}$
= $\frac{102}{103}$
c) $\frac{8}{9}$ - $\frac{1}{72}$ - $\frac{1}{56}$ - $\frac{1}{42}$ - ... - $\frac{1}{6}$ - $\frac{1}{2}$
=$\frac{8}{9}$ - ( $\frac{8}{9}$ + $\frac{1}{56}$ + $\frac{1}{42}$ + ... + $\frac{1}{6}$ + $\frac{1}{2}$ )
=$\frac{8}{9}$ - ( $\frac{1}{9.8}$ + $\frac{1}{8.7}$ + $\frac{1}{7.6}$ + ... + $\frac{1}{3.2}$ +
$\frac{1}{2.1}$
=$\frac{8}{9}$ - ( 1- $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + ... + $\frac{1}{8}$ - $\frac{1}{9}$ )
= $\frac{8}{9}$ - ( 1- $\frac{1}{9}$ )
= $\frac{8}{9}$ - $\frac{8}{9}$
= 0
