= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101
=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)
=1/2 x (1/3 - 1/99)
=1/2 x (1/3 - 1/101)
=1/2 x (98/303)
=1/15 + 1/35 + 1/63 +1/99+...+1/9999
=49/303
\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)
\(=\frac{98}{303}\)
Ta có :A=1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A=2/3.5+2/5.7+2/7.2/9.12/11.13
2A=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
2A=1/3-1/13
2A=10/39
=> A=
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+....+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{101-3}{303}=\frac{1}{2}.\frac{88}{303}=\frac{44}{303}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+..+\frac{1}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A =\(\frac{98}{303}:2\)
A =\(\frac{49}{303}\)
Các bạn ơi!!ủng hộ mik nhaaaa
