A=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+....+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
tính nhanh \(\dfrac{1}{2}\) +\(\dfrac{1}{4}\) +\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\) +.....+\(\dfrac{1}{512}\) +\(\dfrac{1}{1024}\)
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
S=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
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:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
Tính nhanh
A=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{1024}\right)\)
B=4.5100 .\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\right)+1\)
Cho \(A=1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{10}-...-\dfrac{1}{512}.CM:A< 0,002\)
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)
T=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{512}\right)+\left(1-\dfrac{1}{1024}\right)\)
GIÚP MK VỚI 
\(A=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{512}\right)+\left(1-\dfrac{1}{1024}\right)\)
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(2A=-\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)-\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)
\(A=-2\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
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
