Question

\(y=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)

Tìm y

Answer 1

\(y=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)\(=2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)=2\cdot\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)