1/2+1/4+1/8+......+1/256+1/512+1/1024
giúp mk cói mk đanh cần gấp
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BT
1
16 tháng 4 2020
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)
=> A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^8}\)
=> 2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^7}\)
=> 2A-A=\(\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^8}\right)\)
=> A=\(1-\frac{1}{2^8}\)
NT
6
20 tháng 7 2020
1+2+4+8+......+256+512
Phân tích :
1+1 = 2
2+2 =4
4+4=8
...
256+256 = 512
( Lấy 1 số cộng vs chính nó ! )
Học tốt !
20 tháng 7 2020
Bài làm
1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512
= ( 2 + 8 ) + ( 4 + 16 ) + ( 128 + 32 ) + ( 64 + 256 ) + ( 512 + 1 )
= 10 + 20 + 160 + 320 + 513
= 1023
T
6 tháng 7 2019
#)Giải :
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
6 tháng 7 2019
Lời giải
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
9 tháng 8 2019
\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{512}+\frac{1}{1024}=??????????\)
\(< =>1+\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+...+\frac{1}{2\cdot256}+\frac{1}{2\cdot512}\)
\(< =>1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{256}+\frac{1}{2}-\frac{1}{512}\)
\(< =>1+\frac{1}{1}-\frac{1}{512}\)
\(< =>\frac{1023}{512}\)
chuc ban hoc tot nhe :))
KU
1
VA
18 tháng 7 2016
Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(A=\frac{1}{2^2}-\frac{1}{2^8}\)
\(A=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=\frac{1}{512}=\frac{1}{2^9}\)
\(\Rightarrow\frac{63}{2^8}.x=\frac{1}{2^9}\)
\(\Rightarrow x=\frac{1}{2^9}:\frac{63}{2^8}=\frac{1}{2^9}.\frac{2^8}{63}=\frac{1}{2.63}=\frac{1}{126}\)
Ủng hộ mk nha !!! ^_^
DV
1
7 tháng 7 2016
A=1/2+1/4+1/8.....+1/256+1/512
2A=1+1/2+1/4+1/8...1/256
A=(1+1/2+1/4+1/8...1/256)-(1/2+1/4+1/8.....+1/256+1/512)
A=1-1/512
A=511/512
511/512
PV
1
3 tháng 10 2021
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\\ 2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\\ A=1-\dfrac{1}{2^9}=\dfrac{511}{512}\)
LA
2
16 tháng 8 2017
\(D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Leftrightarrow2D=1+\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{256}\)
\(\Leftrightarrow2D-D=\left(1+\dfrac{1}{2}+.....+\dfrac{1}{256}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{512}\right)\)
\(\Leftrightarrow D=1-\dfrac{1}{512}=\dfrac{511}{512}\)
1-1/1024=1023/1024 tick mình nhé
Đặt `A = 1/2+1/4+......+1/1024`
`2 xx A = 2 xx (1/2+1/4+......+1/1024)`
`2 xx A = 1 + 1/2 + ... + 1/512`
`2 xx A - A = (1 + 1/2 + ... + 1/512)-(1/2+1/4+......+1/1024)`
`A = 1 - 1/1024`
`A = 1023/1024`