\(\frac{511}{1280}\)

511/1280

A = 1/5 + 1/10 + 1/20 + 1/40 + 1/80 +1/160 + 1/320 + 1/640 + 1/1280

A x 2 = 2/5 + 1/5 + 1/10 + 1/20 + 1/40 + 1/80 +1/160 + 1/320 + 1/640 

A x 2 - A = A = 2/5 - 1/1280 = 511/1280

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

Số lượng số của A là : ( 1280-5):5+1=256 (số)

A= (1280+5)x256:2=164480

A có số số hạng là :

(1280 - 5 ):5+1 = 256 ( số hạng )

A = ( 1280 + 5 ) x 256 : 2= 164480

mink nhanh nhất

Số số hạng của dãy số đó là :

( 1280 - 5 ) : 5 + 1 = 256 ( số số hạng )

A = ( 5 + 1280 ) x 256 : 2 = 164480

                                      Đ/s : A = 164480 . 

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

\(A=\dfrac{1}{2^0.5}+\dfrac{1}{2^1.5}+\dfrac{1}{2^2.5}+...+\dfrac{1}{2^8.5}\)

\(5A=\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^8}\)

\(5A=2-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+...++\dfrac{1}{128}+\dfrac{1}{256}\)

\(5A=2-\dfrac{1}{256}=\dfrac{511}{256}\)

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

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