C= 1/5+1/10+1/20+1/40+1/80+...........+1280
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18 tháng 6 2021
A = \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)
= \(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
= \(\dfrac{2}{5}-\dfrac{1}{1280}=\dfrac{511}{1280}\)
19 tháng 6 2021
Giải:
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
\(=\dfrac{2}{5}-\dfrac{1}{1280}\)
\(=\dfrac{511}{1280}\)
18 tháng 8 2016
\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)
\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)
TG
30 tháng 10 2016
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
C = \(\frac{1}{5}\)+\(\frac{1}{10}\)+\(\frac{1}{20}\)+\(\frac{1}{40}\)+\(\frac{1}{80}\)+........+\(\frac{1}{1280}\)
2C = 2 . ( \(\frac{1}{5}\)+\(\frac{1}{10}\)+.......+\(\frac{1}{1280}\))
2C = \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\)
2C-C = ( \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+......+\(\frac{1}{1280}\)) - (\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\))
C . ( 2-1) = \(\frac{2}{5}\)
C = \(\frac{2}{5}\)
Vậy C = \(\frac{2}{5}\)
\(C=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+........+\frac{1}{1280}\)
\(\Rightarrow2C=2\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow2C=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+.............+\frac{1}{1280}\)
\(\Rightarrow2C-C=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+............+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow C=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{512}{1280}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{511}{1280}\)
Vậy C = \(\frac{511}{1280}\)