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a/ Đặt 1/5= a, ta có:
1/5 + 1/10 + 1/20 + 1/40 + ... + 1/1280
= 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
A = 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
2 x A = 2/a + 1/a + 1/2 x a + 1/4 x a + ... + 1/128 x a
=> A = 2/a - 1/256 x a = 2/5 - 1/1280 = 511/1280
b/
\(\frac{121}{27}.\frac{54}{11}=\frac{11.11.27.2}{27.11}=11.2=22\)
\(\frac{100}{21}:\frac{25}{126}=\frac{100}{21}.\frac{126}{25}=\frac{25.4.21.6}{21.25}=4.6=24\)
=> \(22< n< 24\)
=> \(n=23\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(A.2=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\)
\(A.2-A=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\right)\)
\(A=\frac{2}{5}-\frac{1}{1280}=\frac{511}{1280}\)
b) \(\frac{121}{27}.\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)
\(22< n< 24\)
=> n = 23
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)
= \(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
= \(\dfrac{2}{5}-\dfrac{1}{1280}=\dfrac{511}{1280}\)
Giải:
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
\(=\dfrac{2}{5}-\dfrac{1}{1280}\)
\(=\dfrac{511}{1280}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/5 + 1/5 - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280
= 1/5 + 1/5 - 1/1280 = 511/1280
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: \(\frac{121}{27}.\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)
\(\Leftrightarrow22< n< 24\)
\(\Rightarrow n=23\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{40}\)+ .......+ \(\dfrac{1}{640}\)+ \(\dfrac{1}{1280}\)
2A = \(\dfrac{2}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{10}\) + \(\dfrac{1}{20}\)+ \(\dfrac{1}{40}\)+.........+ \(\dfrac{1}{640}\)
2A - A = \(\dfrac{2}{5}\)- \(\dfrac{1}{1280}\)
A = \(\dfrac{512}{1280}\) - \(\dfrac{1}{1280}\)
A = \(\dfrac{511}{1280}\)