\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

Dễ thôi bạn à

\(A=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)

\(A=\frac{1.3+1}{1.3}+\frac{3.5+1}{3.5}+\frac{5.7+1}{5.7}+...+\frac{49.50+1}{49.51}\)

\(A=\frac{1.3}{1.3}+\frac{1}{1.3}+\frac{3.5}{3.5}+\frac{1}{3.5}+\frac{5.7}{5.7}+\frac{1}{5.7}+...+\frac{49.51}{49.51}+\frac{1}{49.51}\)

\(A=1+\frac{1}{1.3}+1+\frac{1}{3.5}+1+\frac{1}{5.7}+...+1+\frac{1}{49.51}\) (có: (51 - 3) : 2 + 1 = 25 chữ số 1)

\(A=25+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{49}-\frac{1}{51}\right)\)

\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=25+\frac{1}{2}.\left(1-\frac{1}{51}\right)\)

\(A=25+\frac{1}{2}.\frac{50}{51}\)

\(A=25+\frac{25}{51}\)

\(A=\frac{1300}{51}\)

thank you

\(S=\frac{4}{1\times3}+\frac{16}{3\times5}+\frac{36}{5\times7}+...+\frac{2500}{49\times51}\)

\(=\frac{1\times3+1}{1\times3}+\frac{3\times5+1}{3\times5}+\frac{5\times7+1}{5\times7}+...+\frac{49\times51+1}{49\times51}\)

\(=\frac{1\times3}{1\times3}+\frac{1}{1\times3}+\frac{3\times5}{3\times5}+\frac{1}{3\times5}+\frac{5\times7}{5\times7}+\frac{1}{5\times7}+...+\frac{49\times51}{49\times51}+\frac{1}{49\times51}\)

\(=1+\frac{1}{1\times3}+1+\frac{1}{3\times5}+1+\frac{1}{5\times7}+...+\frac{1}{49\times51}\) (  Có : \(\left(51-3\right)\div2+1=25\)chữ số 1 )

\(=25+\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{49\times51}\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\frac{50}{51}\)

\(=25+\frac{25}{51}\)

\(=\frac{1300}{51}\)

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)

\(=\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+...+\frac{2500}{2499}\)

\(=1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+...+1+\frac{1}{2499}\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2500}\right)\)

\(=25+\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=1-\frac{1}{51}=\frac{50}{51}\)

\(\Rightarrow S=25+\frac{50}{51}=\frac{1325}{51}\)

Vậy S=\(\frac{1325}{51}\)

Ta có:

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+........+\frac{2500}{49.51}\)

Đặt \(S=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(\Rightarrow S=\frac{2}{2}.\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\right)\)

\(\Rightarrow S=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{3}{99.101}\right)\)

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

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

\(\Rightarrow S=\frac{3}{2}.\frac{100}{101}\)

\(\Rightarrow S=\frac{150}{101}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}\)

\(=\frac{1009}{673}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)