Question

Tính nhanh A = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+19}\)

Answer 1

A = \(1+\frac{1}{\left(2\times3\right):2}+\frac{1}{\left(3\times4\right):2}+....+\frac{1}{\left(19\times20\right):2}\)

A = \(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{19\times20}\)

A = \(2\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)=2\times\left(\frac{1}{1}-\frac{1}{20}\right)=\frac{19}{10}\)

Answer 2

Bài này dễ ẹc mà !      

Answer 3

\(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+19}=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{19.20}\)

 

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}\)