TK

\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+.......+\dfrac{1}{1280}\)

\(=\dfrac{1}{5}+\dfrac{1}{5.2}+\dfrac{1}{5.4}+\dfrac{1}{5.8}+.....+\dfrac{1}{5.256}\)

\(=\dfrac{1}{5}+\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^8}\right)\)

Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^8}\)

\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^7}\)

\(2A-A=2-\dfrac{1}{2^3}\)

Thay A vào ta được :

\(\dfrac{1}{5}.A=\dfrac{1}{5}.\left(2-\dfrac{1}{2^8}\right)=\dfrac{511}{1280}\)

\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)

\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)

\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)

\(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)

\(=\frac{1}{5}\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

Đặt \(A=1+\frac{1}{2}+...+\frac{1}{2^8}\)

\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

\(2A=2+1+...+\frac{1}{2^7}\)

\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

\(A=2-\frac{1}{2^8}\).Thay A vào đc: \(\frac{1}{5}\cdot\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)

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 = \(\frac{511}{1280}\)

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

ok bạn

1/5+1/10+1/20+...+1/1280

=1/1x5+1/

B=51​+101​+201​+401​+...+12801​

�=1⋅15+12⋅15+14⋅15+18⋅15+...+1256⋅15B=1⋅51​+21​⋅51​+41​⋅51​+81​⋅51​+...+2561​⋅51​

�=15⋅(1+12+14+18+...+1256)B=51​⋅(1+21​+41​+81​+...+2561​)

Đặt �=1+12+14+18+...+1256A=1+21​+41​+81​+...+2561​

⇒2�=2+1+12+14+...+1128⇒2A=2+1+21​+41​+...+1281​

⇒2�−�=2−1256⇒2A−A=2−2561​

�=2−1256A=2−2561​

Thay A vào B

có: �=15.(2−1256)=15⋅511256=5111280B=51​.(2−2561​)=51​⋅256511​=1280511​
 

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