Giải :
\(N=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
=> \(N=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)\right)\)
=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\right)\)
=> \(N=\frac{98}{303}\)
N=1/2x(1/3-1/5+1/5-1/7+.....+1/99-1/101)
N=1/2x(1/3-1/101)
N=1/2x98/303
N=49/303
Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức ta có :
\(\frac{2}{3.5}=\frac{2}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\frac{2}{5.7}=\frac{2}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)\)
........................................
\(\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow\)\(N=1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow\)\(N=\frac{1}{3}-\frac{1}{101}\)
\(\Rightarrow N=\frac{98}{303}\)
Ai thấy đúng thì ủng hộ nha !!!!
\(N=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(N=2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(N=2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(N=\frac{1}{3}-\frac{1}{101}\)
\(N=\frac{98}{303}\)
\(N=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\)\(+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{101}\)
N = 2 * 2 * ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ....+ 1/99 - 1/101 )
N = 2 * 2 * ( 1/3 - 1/101 )
N = 4 * 98/303
N = 392/303
N= 2/3.5+2/5.7+2/7.9+...+2/99.101
N=1/3_1/5+1/5_1/7+1/7_1/9+...+
1/99_1/101
N=1/3_1/101=101/303_3/303=98/303
N= 2 * 2 * ( 1/3- 1/5+1/5-1/7 + 1/7 - 1/9 + .........+ 1/99 - 1/101)
N=2*2 * (1/3 - 1/101)
N= 4*98/ 303
N=392/303
Bạn nhé! k mik đi
\(N=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
=> \(N=\frac{1}{3}-\frac{1}{5} +\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=> \(N=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
Vậy \(N=\frac{98}{303}\)