Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)

<=>A=1-1/1024

<=>A=1023/1024

Vậy biểu thức đã cho = 1023/1024

a: Đặt \(A=\frac12-\frac14+\frac18-\frac{1}{16}+\cdots-\frac{1}{1024}\)

=>\(A=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(2A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}\)

=>\(2A+A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^9}+\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{10}}\)

=>\(3A=1-\frac{1}{2^{10}}<1\)

=>\(A<\frac13\)

b: Đặt \(B=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>\(3B+B=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>\(3A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>\(3A+A=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

=>\(4A=-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4B=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}=\frac34-\frac{403}{4\cdot3^{100}}\)

=>\(4B<\frac34\)

=>\(B<\frac{3}{16}\)

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

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

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

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

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

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

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

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

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

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

Lời giải:

Sửa lại đề:

\(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{512}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

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

Trừ theo vế:

\(A=2A-A=\frac{1}{2^9}< 0,002\) (đpcm)

a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{197.199}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(=\dfrac{1}{3}-\dfrac{1}{199}=\dfrac{196}{597}\)

b, \(B=1+2+4+...+1024\)

\(\Rightarrow2B=2+4+8+...+2056\)

\(\Rightarrow2B-B=\left(2+4+8+...+2056\right)-\left(1+2+4+...+1024\right)\)

\(\Rightarrow B=2056-1=2055\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{199}\)

\(A=\dfrac{199}{597}-\dfrac{3}{597}=\dfrac{196}{597}\)

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{197.199}\)

\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{199-197}{197.199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{197}-\dfrac{1}{199}\)

\(A=\dfrac{1}{3}-\dfrac{1}{999}\)

\(A=\dfrac{196}{697}\)

\(B=1+2+4+8+16+...+512+1024\)

\(2B=2+4+8+32+...+1024+2048\)

\(B=\left(2+4+8+...+2048\right)-\left(1+2+4+...+1024\right)\)

\(B=2048-1\)

\(B=2047\)

S=1/2+1/4+1/8+...+1/1024=(1/2)^1+(1/2)^2+(1/2)^3+...+(1/2)^10

2S=1+(1/2)^1+(1/2)^2+...+(1/2)^9

2S-S=1-(1/2)^10

vậy S=1-(1/2)^10

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

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

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

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

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

\(A=-2\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)

\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Đặt ⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256

⇒A=1−1512=511512