Question

\(\frac{1}{3.4}+\frac{1}{4.13}+\frac{1}{13.9}+\frac{1}{9.23}+.........+\frac{1}{49.103}\\ \)

Answer 1

Đúng mik cho 3 tk

Answer 2

   \(\frac{1}{3\cdot4}+\frac{1}{4.13}+\frac{1}{13.9}+\frac{1}{9\cdot23}+...+\frac{1}{49\cdot103}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{13}+\frac{1}{13}-\frac{1}{9}+\frac{1}{9}-\frac{1}{23}+...+\frac{1}{49}-\frac{1}{103}\)

= \(\frac{1}{3}-\frac{1}{103}\)

= \(\frac{100}{309}\)

Answer 3

\(\frac{1}{3\times4}+\frac{1}{4\times13}+\frac{1}{13\times9}+...+\frac{1}{49\times103}\)

\(=\frac{2}{3\times8}+\frac{2}{8\times13}+\frac{2}{13\times18}+...+\frac{2}{98\times103}\)

\(=\frac{2}{5}\times\left(\frac{5}{3\times8}+\frac{5}{8\times13}+\frac{5}{13\times18}+...+\frac{5}{98\times103}\right)\)

\(=\frac{2}{5}\times\left(\frac{8-3}{3\times8}+\frac{13-8}{8\times13}+\frac{18-13}{13\times18}+...+\frac{103-98}{98\times103}\right)\)

\(=\frac{2}{5}\times\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{98}-\frac{1}{103}\right)\)

\(=\frac{2}{5}\times\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(=\frac{40}{309}\)