Question

 tính A =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...\frac{1}{48.49.50}\)

 

Answer 1

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2450}\right)\)

\(A=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)

Answer 2

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{1.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}\right)-\frac{1}{3.4}+...\frac{1}{2}\left(\frac{1}{48.49}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(A=\left(\frac{1}{2}.\frac{1}{49.50}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{2}.\frac{1}{1225}=\frac{306}{1225}\)