Question

Tính nhanh tổng\(B=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+8}\)

Answer 1

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+8}\)

\(=\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+...+\frac{1}{\frac{8\cdot9}{2}}=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)=2\left(\frac{1}{2}-\frac{1}{9}\right)=2\cdot\frac{9-2}{18}=\frac{18-4}{18}=\frac{14}{18}=\frac{7}{9}\)