Đặt \(M=68+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}\)
\(\Rightarrow A\times\frac{1}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{380}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
\(\Rightarrow M=68+\frac{19}{20}=\frac{1379}{20}\)
Vậy \(68+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{190}=\frac{1379}{20}\)
Cho tớ sửa lại dòng thứ 2 từ dưới lên :
\(\Rightarrow A=\frac{19}{20}:\frac{1}{2}=\frac{19}{10}\)
\(\Rightarrow M=68+\frac{19}{10}=\frac{699}{10}\)
Vậy ...
