Question

Tính nhanh:

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}\)

Answer 1

A = 1/4 + 1/28 + 1/70 +...+ 1/9700

A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100

3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100

3A = 1 - 1/100

3A = 99/100

A=99/100:3=33/100

Answer 2

\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

Answer 3

= \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\)

= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

= \(\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{100}\right)\)

= \(\frac{1}{3}.\frac{99}{100}\)

=\(\frac{33}{100}\)