Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
tìm số nguyen x: \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)
Gọi \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{x}=\dfrac{1023}{1024}\)
VẬy x là một lũy thừa của 2. Đặt x = 2y , ta có:
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^y}\)
\(\Rightarrow2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}\)
\(\Rightarrow2A-A=1+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{y-1}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^8}\right)\)
\(=A-\dfrac{1}{2^y}\)
Vậy \(1-\dfrac{1}{2^y}=\dfrac{1023}{1024}\Leftrightarrow\dfrac{1}{2^y}=\dfrac{1}{1024}\Leftrightarrow2^y=1024\Rightarrow x=1024\)
Vậy x = 1024
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=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
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
A=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+....+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
\(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\) có:
\(2A=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{1}{2}-\left(\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{1024}=\dfrac{1}{1024}\)
Vậy...
Cách của Tuấn Anh Phan Nguyễn đây.
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\right]\)
\(=\dfrac{1}{2}-\left[\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{512}-\dfrac{1}{1024}\right)\right]\)\(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{1024}\right)=\dfrac{1}{1024}.\)
Tách nó ra thành 21;22;23..........đến 210đó bạn
-1-\(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
Đặt :
\(H=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-..........-\dfrac{1}{1024}\)
\(\Leftrightarrow H=-1-\left(\dfrac{1}{2}+\dfrac{1}{4}+...........+\dfrac{1}{1024}\right)\)
Đặt :
\(T=\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{1024}\)
\(\Leftrightarrow T=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2T=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\)
\(\Leftrightarrow2T-T=\left(1+\dfrac{1}{2}+.....+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow T=1-\dfrac{1}{2^{10}}\)
\(\Leftrightarrow H=-1-\left(1-\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow H=-1-1+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow H=-2+\dfrac{1}{2^{10}}\)
Đặt \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)
\(A=-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
\(2B-B=1-\dfrac{1}{1024}\)
\(\Rightarrow B=\dfrac{1023}{1024}\)
\(\Rightarrow A=-\dfrac{1023}{1024}\)
Đặt
\(W=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-....-\dfrac{1}{1024}\)
\(W=-1-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\right)\)
\(W=-1-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)
Đặt:
\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\)
\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)
\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}\)
Thay \(A\) vào \(W\) ta có:
\(W=-1-1+\dfrac{1}{2^{10}}=-2+\dfrac{1}{2^{10}}\)
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\). mọi ng cho mình lời giải cụ thể với ạ.
M = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
4.M = 1 + 1/4 + 1/16 + 1/64 + 1/256
4M - M = (1 + 1/4 + 1/16 + 1/64 + 1/256 ) - ( 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 )
3M = 1 - 1/1024
3M = 1023/1024
M = 341/1024
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\)
=\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\)
=>4M=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)
=>4M-M=3M=(1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\))=1-\(\dfrac{1}{4^5}\)=\(\dfrac{1023}{1024}\)
=>M=\(\dfrac{1023}{1024}\):3=\(\dfrac{341}{1024}\)
`@` `\text {Ans}`
`\downarrow`
\(M=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\)
\(\Rightarrow4M=1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\)
\(\Rightarrow4M-M=\) \(\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)-\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\right)\)
\(=\) \(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256}-\dfrac{1}{1024}\)
\(=1-\dfrac{1}{1024}\)
\(=\dfrac{1024}{1024}-\dfrac{1}{1024}=\dfrac{1023}{1024}\)
`4M - M = 3M`
\(\Rightarrow3M=\dfrac{1023}{1024}\)
\(\Rightarrow M=\dfrac{1023}{1024}\div3\)
\(\Rightarrow M=\dfrac{341}{1024}\)
Vậy, `M = `\(\dfrac{341}{1024}\)