1/3x5 +1/5x7+1/7x9 +1/9x11+...+1/99x101
1/3-1/5+1/5-1/7+...+1/99-1/101
1/3-1/101
98/303
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.9}\)
\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{111.9}\right)\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{1}{111.9}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{111}-\frac{1}{9}\)
\(=\frac{1}{3}-\frac{1}{9}\)
\(=\frac{2}{9}\)
Ta có :
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+..+\frac{1}{999}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{3}-\frac{1}{101}\)
\(A=\frac{98}{303}:2\)
\(A=\frac{49}{303}\)
Ủng hộ mk nha !!! ^_^
