Question

tính:  \(\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)

Answer 1

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)

\(=\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{50.51:2}=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{100}\right)=2.\frac{49}{100}=\frac{49}{50}\)

Answer 2

=\(\frac{49}{50}\)