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S
4 tháng 3 2016
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
HW
1
13 tháng 3 2017
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
26 tháng 7 2016
\(M=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{502}+\frac{1}{1024}\)
\(M\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{502}+\frac{1}{1024}\right)\cdot2\)
\(M\cdot2=\frac{1}{2}\cdot2+\frac{1}{4}\cdot2+\frac{1}{8}\cdot2+\frac{1}{16}\cdot2+...+\frac{1}{502}\cdot2+\frac{1}{1024}\cdot2\)
\(M\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{504}\)
\(M\cdot2-M=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{502}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{502}+\frac{1}{1024}\right)\)
\(M=1-\frac{1}{1024}\)
\(M=\frac{1023}{1024}\)
12 tháng 8 2016
Đặt A = 1/2 + 1/4 + 1/8 + ... + 1/1024
2A = 1 + 1/2 + 1/4 + ... + 1/512
2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)
A = 1 - 1/1024
A = 1023/1024
BA
19 tháng 6 2019
B x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
B x 2 = 1 - 1/1024 + A
B x 2 - B = 1 - 1/1024
B = 1 - 1/1024
B = 1023 /1024
BA
19 tháng 6 2019
làm lại bài kia sai 1 chỗ
\(B=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-...+\frac{1}{1024}\)
\(\Rightarrow Bx2=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\frac{1}{1024}\)
\(\Rightarrow Bx2=1-\frac{1}{1024}+B\)
\(\Rightarrow Bx2-B=1-\frac{1}{1024}\)
\(\Rightarrow B=\frac{1023}{1024}\)
DL
1
TG
17 tháng 2 2022
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DD
1
TN
2
7 tháng 6 2018
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
LD
7 tháng 6 2018
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
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: ...
$Toru$