\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=\(1-\frac{1}{101}=\frac{100}{101}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(2\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(2\left(1-\frac{1}{101}\right)\)
\(2\times\frac{100}{101}\)
\(\frac{200}{101}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
= 2(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\))
= 2(\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\))
= 2(\(\frac{1}{1}-\frac{1}{101}\))
= .\(2.\frac{100}{101}\)
= \(\frac{50}{101}\)
đặt
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\)
chúc bạn hok tốt!
