\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}\div2=\frac{49}{303}\)
*Lưu ý là dấu "." là nhân nhé! Nếu bạn không chắc thì từ cái khúc 1/3 - 1/101 bạn tự làm, xong rồi chia 2 nha!
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
A = \(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}+.....+\frac{1}{99x101}\)
A x 2 = \(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+.....+\frac{2}{99x101}\)
A x 2 = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+.....+\frac{1}{99}-\frac{1}{101}\)
A x 2 = \(\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
A = \(\frac{98}{303}:2=\frac{49}{303}\)
Vậy A = 49/303
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3\text{ x }5}+\frac{1}{5\text{ x }7}+\frac{1}{7\text{ x }9}+\frac{1}{9\text{ x }11}+.....+\frac{1}{99\text{ x }101}\)
\(\Rightarrow2A=\frac{2}{3\text{ x }5}+\frac{2}{\text{5 x 7 }}+\frac{2}{7\text{ x }9}+....+\frac{2}{99\text{ x }100}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+.....+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
\(A=\frac{98}{303}:2=\frac{49}{303}\)
