A=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
A=\(\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
A=\(\frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\right]\)
A=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
A=\(\frac{1}{2}\cdot\frac{98}{303}\)
A=\(\frac{49}{303}\)
\(\cdot\) là dấu nhân
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3X5}+\frac{1}{5X7}+\frac{1}{7X9}+\frac{1}{9X11}+...+\frac{1}{99X101}\)
\(2A=\frac{2}{3X5}+\frac{2}{5X7}+\frac{2}{7X9}+\frac{2}{9X11}+...+\frac{2}{99X101}\)
\(=\frac{5-3}{3X5}+\frac{7-5}{5X7}+\frac{9-7}{7X9}+\frac{11-9}{9X11}+...+\frac{101-99}{99X101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
Vậy A là : \(\frac{98}{303}:2=\frac{49}{303}\)
=>A=1/3.5+1/5.7+...+1/99.101
A=1/2.(2/3.5+2/5.7+...+2/99.101)
A=1/2.{[1/3-1/5]+[1/5-1/7]+...+[1/99-1/101]}
A=1/2.{1/3-1/101}
A=1/2.98/303
A=49/303
