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
\(A=\frac{1}{5}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3X5}+\frac{1}{5X7}+\frac{1}{7X9}+\frac{1}{9X11}+...+\frac{1}{99X101}\)
\(2A=\frac{2}{3X5}+\frac{2}{5X7}+\frac{2}{7X9}+\frac{2}{9X11}+...+\frac{2}{99X101}\)
\(=\frac{5-3}{3X5}+\frac{7-5}{7X9}+\frac{9-7}{9X7}+\frac{11-9}{9X11}+...+\frac{101-99}{101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
Từ đó ta suy ra:\(A=\frac{98}{303}:2=\frac{49}{303}\)
