Question

Thực hiện phép tính:

\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\frac{1-3-5-7-..-49}{89}\)

Answer 1

\(=\left[\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right)+\frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right)+\frac{1}{5}\left(\frac{1}{14}-\frac{1}{19}\right)+...+\frac{1}{5}\left(\frac{1}{44}-\frac{1}{49}\right)\right]\cdot\frac{1-\left(3+5+...+49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(52+52+...+52\right)\left\{12\text{ số 52}\right\}}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)

\(=\frac{9}{196}\cdot-7=\frac{9}{28}\)