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​
 

bfvcf

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:

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

\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\) 

\(=\dfrac{2}{5}-\dfrac{1}{1280}\) 

\(=\dfrac{511}{1280}\)

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: \(\frac{1}{5}\cdot\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)

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/ 

\(\frac{121}{27}.\frac{54}{11}=\frac{11.11.27.2}{27.11}=11.2=22\)

\(\frac{100}{21}:\frac{25}{126}=\frac{100}{21}.\frac{126}{25}=\frac{25.4.21.6}{21.25}=4.6=24\)

=> \(22< n< 24\)

=> \(n=23\)

KO HIEU GIO TAY