Question

[đức tiến đạt]

A=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.......+\(\frac{2}{13.15}\)+\(\frac{2}{15.7}\)

B=\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+.......+\(\frac{4}{9.11}\)+\(\frac{4}{11.13}\)

Answer 1

[Mạnh Lê]

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{17}\)

\(A=1-\frac{1}{17}\)

\(A=\frac{16}{17}\)

\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)

\(B=\frac{4}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(B=\frac{4}{2}\left(1-\frac{1}{13}\right)\)

\(B=\frac{4}{2}\cdot\frac{12}{13}\)

\(B=\frac{24}{13}\)

Answer 2

[Nguyễn Huy Quang]

=> A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)

=> A= \(\frac{1}{1}-\frac{1}{17}\)

=> A= \(\frac{16}{17}\)

\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{13}\right)\)

\(\Rightarrow B=2.\frac{12}{13}\)

\(\Rightarrow B=\frac{24}{13}\)