Question

Tính các tổng:

M = \(\dfrac{6}{2\cdot5}+\dfrac{6}{5\cdot8}+\dfrac{6}{8\cdot11}+...+\dfrac{6}{47\cdot50}\)

K = \(\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}+...+\dfrac{1}{43\cdot45}\)

Answer 1

\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)

\(\Rightarrow\dfrac{M}{2}=\dfrac{6:2}{2.5}+...+\dfrac{6:2}{47.50}\)

\(=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\)

\(=\dfrac{1}{2}-\dfrac{1}{50}\)

\(=\dfrac{12}{25}\)

\(\Rightarrow M=\dfrac{12}{25}.2=\dfrac{24}{25}\)

\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)

\(\Rightarrow2K=\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\)

\(=\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\)

\(=\dfrac{1}{9}-\dfrac{1}{45}\)

\(=\dfrac{4}{45}\)

\(\Rightarrow K=\dfrac{4}{45}:2=\dfrac{2}{45}\)

Answer 2

\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)

\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{5}+\dfrac{6}{5}-\dfrac{6}{8}+\dfrac{6}{8}-\dfrac{6}{11}+...+\dfrac{6}{47}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\left(\dfrac{150}{50}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\dfrac{144}{50}\)

\(M=\dfrac{144}{25}\)

\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)

\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\left(\dfrac{5}{45}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\dfrac{4}{45}\)

\(K=\dfrac{2}{45}\)