Question

tính :

B = \(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...............+\frac{2}{399}\)

Answer 1

Ta có \(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

\(=2.\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{2}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{2}{7}\)

Vậy \(B=\frac{2}{7}\)