Question

Tính giá trị biểu thức : 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+50)

 

 

 

 

Answer 1

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

\(=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2550}\)

\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{50\cdot51}\)

\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{50\cdot51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)