Question

Tính nhanh:

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\)

Answer 1

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)

\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{101}\right)\)

Tự tính 

Answer 2

  \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\)

    \(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)

   \(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)

   \(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)

    \(=2\left(\frac{1}{2}-\frac{1}{101}\right)\)

    \(=2.\frac{99}{202}\)

     \(=\frac{99}{101}\)