Lời giải:
$A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}$
$3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{197.200}$
$3A=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}$
$=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}$
$=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}$
$A=\frac{13}{200}$
Ta có: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{197\cdot200}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{39}{200}=\dfrac{13}{200}\)
