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

A=1-1/3+1/3-1/5+....+1/99-1/100

A=1-1/100=99/100

200A= 99/100   .200=198

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

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

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

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

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

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

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

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

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

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

1-1/100 , ủng hộ mk nha

=>2S=2/1.3+2/3.5+....+2/99.100

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