Question

[Cùng học Toán]

Tính nhanh:

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+......+\frac{1}{1+2+3+....+10}\)

Answer 1

[Bùi Anh Tuấn]

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+10}\)

\(=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\cdot\frac{9}{22}\)

\(=\frac{9}{11}\)