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Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}-\frac{1}{1024}\)
\(\Rightarrow2A=-2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
\(\Rightarrow2A-A=-2+\frac{1}{1024}\)
\(A=-2+\frac{1}{1024}\)
= -1-1+1/2-1/2+1/4-1/4+1/8-...+1/512-1/1024
=-1-1-1/1024
=-2\(\dfrac{1}{1024}\)
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(-2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(-2A+A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(-A=2-\frac{1}{1024}\)
\(A=\frac{1}{1024}-2\)
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(2A=-\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}-2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)
\(A=-2+\frac{1}{2^{10}}\)
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