A = 1/5 + 1/10 + 1/20 + 1/40 + ..... + 1/1280

A x 2 = 2/5 - ( 1 /5 + 1/10 + 1/20 + 1/40 + ... + 1/1280 ) - 1/1280

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

A x 2 - A = 2/5 - 1/1280

A  = 2/5 - 1/1280

A = 511/1280

A = 1/5 + 1/10 + 1/20 + 1/40 + ..... + 1/1280

A x 2 = 2/5 - ( 1 /5 + 1/10 + 1/20 + 1/40 + ... + 1/1280 ) - 1/1280

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

A x 2 - A = 2/5 - 1/1280

A  = 2/5 - 1/1280

A = 511/1280

\(=\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{2}+\frac{1}{5}\cdot\frac{1}{2^2}+\frac{1}{5}\cdot\frac{1}{2^3}+...+\frac{1}{5}\cdot\frac{1}{2^8}\)

\(=\frac{1}{5}\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)=\frac{1}{5}\cdot2\cdot\left(1-\frac{1}{2}\right)\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(=\frac{2}{5}\cdot\left(1-\frac{1}{2^9}\right)=\frac{2\cdot\left(2^9-1\right)}{5\cdot2^9}=\frac{511}{1280}\)

1/5+1/10+1/20+1/40+...+1/1280= 
=1/5(1+1/2^1+1/2^2+...+1/2^8) 
=1/5*(1-1/2^9)/(1-1/2) 
=2/5*(1-1/2^9)

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

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

Gọi tổng trên là A

Ta có : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{2560}\)

\(2A=2\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(2A=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)

\(2A-A=\left(\frac{2}{5}+\frac{1}{5}+...+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(A\left(2-1\right)=\frac{2}{5}-\frac{1}{2560}\)

\(A.1=\frac{1024}{2560}-\frac{1}{2560}\) 

\(A=\frac{1023}{2560}\)

Ta có : A = 1/5 + 1/10 + 1/20 + ... + 1/2560

2A = 2 ( 1/5 + 1/10 + ... + 1/2560 )

2A = 2/5 + 1/5 + 1/10 + .. + 1/2560

2A - A = ( 2/5 + 1/5 + ... + 1/1280 ) - ( 1/5 + 1/10 + ... + 1/2560 )

A =  2 - 1 = 2/5 - 1/2560

A.1 = 1024/2560 - 1/2560

A = 1023 = 2560

đáp số sai

tuyên truyền ae chửi báo cáo sai phạm con girl yêu

đừng để nó thôi miên

rất cảm ơn ! gà con ; 1234 ; ...

họ là người tốt đừng để thôi miên 

Bạn lấy phân số 1/1280 nhân cho 2 đi

bạn cứ lấy Ax2 là ra ngay í mà

tui cung dang hoi bai do ?

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 :

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