có dạng này nhưng là số chẵn nhân chãn

2S=2/1.3+2/3.5+....+2/99.101

2S=1-1/3+1/3-1/5+....+1/99-1/101

2S=1-1/101

2S+1/101=1-1/101+1/101=1

Nho tick nha

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(S=1-\frac{1}{101}=\frac{100}{101}\)

\(2S+\frac{1}{101}=\frac{100}{101}\)

\(S=2.\frac{100}{101}+\frac{1}{101}\)

\(\Rightarrow S=\frac{201}{101}\)

****

2S + \(\frac{1}{101}\)=\(\frac{201}{101}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}\Rightarrow2S+\frac{1}{101}=1\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.........+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}\)

\(2S+\frac{1}{101}=1-\frac{1}{101}+\frac{1}{101}=1\)

Nguyễn Nhật Minh đúng rồi

2S+\(\frac{1}{100}\)=1

Cho minh vai li-ke cho tron 130 nha

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}=\frac{100}{101}\)

\(S=\frac{50}{101}\)

còn cần không bạn, mk làm cho

Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{50}{101}\)

suy ra: \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{1}-\frac{1}{x+2}=\frac{50}{101}:\frac{1}{2}=\frac{100}{101}\)

\(\frac{1}{x+2}=1-\frac{100}{101}=\frac{1}{101}\)

suy ra: \(x+2=101\)

suy ra: \(101-2=99\)

S= 1/1.3 + 1/3.5 + 1/5.7 +................+ 1/200.202

=>S=1/2.(2/1.3+2/3.5+2/5.7+...+2/200.202)

=>S=1/2.(3-1/1.3+5-3/3.5+...+202-200/200.202)

=>S=1/2.(1-1/3+1/3-1/5+...+1/200-1/202)

=>S=1/2.(1-1/202)

=>S=1/2.201/202

=>S=201/404

Vậy S=201/404

\(S=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}\times\frac{100}{101}=\frac{50}{101}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.100}\)

\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(S=1-\frac{1}{100}\)

\(S=\frac{99}{100}\)

\(S=\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{99\times101}\) chứ bạn