1/2+1/4+1/8+1/16+...+1/1024
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512 + 1/1024
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512
A x 2 - A = 1 + 1/2 - 1/2+ 1/4 -1/4 + 1/8 -1/8 + 1/16 -1/16 + ... + 1/512 - 1/512 - 1/1024
A = 1 - 1/1024
A = 1023/1024
lớp 1 kinh thế. lớp 6 chứ
chuyển thành công đi cho dẽ nhìn
A=1-(1/2+1/4+1/8+1/16...+1/1024)
B=(1-A)=1/2+1/4+1/8+...+1/1024
B=1/2+1/2^2+1/2^3+..+1/2^10 (cái này có công thức cấp số nhân)
có thể chưa học cách làm bình thừng như sau
1/2.B=1/2^2+1/2^3+...+1/2^10+1/2^11
Trừ cho nhau
1/2B=1/2 -1/2^11 ( cái giữa triệt tiêu)
B=1-1/2^10
A=1-B=1/2^10
DS: 1/2^10
gọi A=1/2+1/4+1/8+...+1/1024
2xA=1+1/2+1/4+.....+1/512
2xA-A=(1+1/2+1/4+....+1/512)-(1/2+1/4+1/8+...+1/1024)
A=1-1/1024
=1023/1024
vậy A=1023/1024
A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)
=1-1/1024
=1023/1024
A = 1/2+1/4+...+1/2048
2A= 1+ 1/2+ 1/4+...+1/1024
2A-A= ( 1+ 1/2+...+1/1024 ) - (1/2+1/4+...+2048)
A= 1- 1/2048
A= 2047/2048
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: ...
$Toru$