*Tham khảo 

 A = 1/2 + 1/4 + 1/8 + ... + 1/1024

2A = 1 + 1/2 + 1/4 + ... + 1/512

2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)

A = 1 - 1/1024

A = 1023/1024

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)

\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)

\(A=\frac{1023}{1024}\)

A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

2 x A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)

2 x A - A = \(\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)

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

A=\(\frac{1023}{1024}\)

1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1 – 1/2 + 1/2- 1/4 + 1/4 – 1/8 + 1/8 – 1/16 + 1/16 – 1/32 + 1/32 – 1/64 + 1/64 – 1/128 + 1/128 – 1/256 – 1/256 – 1/512
= 1 – 1/512

= 511/512 .

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Cảm ơn các bạn

Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128) -                                     (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)

=> A = 1 - 1/256

=> A = 255/256

Vậy: ...

A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)

= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

=>A = 2A-A =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256

=1-1/256

=255/256

TGV.Quỷ đúng rr đó

giải tri tiết nha

1/5+45/9+1/2+1/3+1/2+1/9+1/15+1/99= ai trả lời đc đưa số tài khoản mik cho 100k

Bạn chỉ cần ghép sao cho số đó tròn chục là đc

Ta có : \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)

\(\Rightarrow2A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}+\frac{1}{2^7}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\right)\)

\(\Rightarrow A=1-\frac{2}{8}=\frac{256}{256}-\frac{1}{256}=\frac{255}{256}\)

Đặt Z =   \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

Ta có: \(Z=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(2Z=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(2Z=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)

\(Z=1-\frac{1}{256}\)

\(Z=\frac{255}{256}\)

ủng hộ lên 1010 điểm nha các bạn

\(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(2\times E=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)

\(2\times E-E=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)\)

\(E=1-\dfrac{1}{256}\)

\(E=\dfrac{256}{256}-\dfrac{1}{256}\)

\(E=\dfrac{255}{256}\)