Question

tính

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

Answer 1

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

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

= \(\frac{1}{1.2}-\frac{1}{49.50}\)

= \(\frac{1}{2}-\frac{1}{2450}\)

= \(\frac{612}{1225}\)

Answer 2

đặt

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

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)

\(\Rightarrow A=\frac{306}{1225}\)

Answer 3

=>2A  = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)

=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49-50}\)

= \(\frac{1}{1.2}-\frac{1}{49.50}\)= \(\frac{1}{2}-\frac{1}{2450}=\frac{612}{1225}\)

Answer 4

sửa nhầm

\(\Rightarrow x=\frac{621}{1225}\)

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