uh đúng rồi

kết quả sai rồi phải là \(\frac{1023}{1024}\)

\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)

\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)

\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$

$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$

$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$

$\frac{3x}{1024}=\frac{1023}{1024}$

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{1024}\) 

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

\(A=-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(2A=-\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}-2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)

\(A=-2+\frac{1}{2^{10}}\)

Ta có: -1 - 1/2 - 1/4 - 1/8 - ... -1/1024

= -1 - (1 - 1/2) - (1/2 - 1/4) - .... - (1/512 - 1/1024)

= -1 - (1 - 1/1024)

= -1 - 1023/1024

= -2047/1024

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

\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{512}\)

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

\(A=2-\dfrac{1}{1024}\)

\(A=\dfrac{2047}{1024}\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\\ =1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\\ \Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{11}}\\ \Rightarrow A-\dfrac{1}{2}A=1-\dfrac{1}{2^{11}}\\ \Rightarrow A=2-\dfrac{1}{2^{10}}\)

Đặt $A=\dfrac12+\dfrac14+\dfrac18+\dfrac{1}{16}+...+\dfrac{1}{1024}$

$A=\dfrac12+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}$

$\dfrac12\cdot A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}}$

$A-\dfrac{1}{2}A=(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}})-(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}})$

$\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{11}}$

$\dfrac{1}{2}A=\dfrac{1}{2}\cdot(1-\dfrac{1}{2^{10}})$

$\Rightarrow A=1-\dfrac{1}{2^{10}}$

Vậy: ...

$Toru$

Đặt A = 1+2+4+8+16+ ... + 1024
2A2+4+8 + 16 +32 + + 2048
2A-A (2 + 4 + 8 + 16 + 32 + ... + 2048) - (1 + 2 + 4 + 8 + 16+..+ 1024)
A = 2048 - 1
A = 2047