\(tinh?-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
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Những câu hỏi liên quan
TV
1
21 tháng 9 2016
Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)
12 tháng 8 2016
\(A=\left(1-\frac{1}{2^1}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^9}\right)+\left(1-\frac{1}{2^{10}}\right)\)
\(A=\left(1+1+1+...+1+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
10 số 1
\(A=10-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(2B-B=1-\frac{1}{2^{10}}=B\)
=> \(A=10-\left(1-\frac{1}{2^{10}}\right)\)
=> \(A=10-1+\frac{1}{2^{10}}\)
=> \(A=9\frac{1}{1024}\)
NY
2
U
15 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
M
15 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}.\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
<=> \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)
<=> \(2A-A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{256}+\frac{1}{512}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}-\frac{1}{1024}\)
<=> \(A=1-\frac{1}{1024}\)
<=> \(A=\frac{1023}{1024}\)
SH
0
NG
0
HN
1
TM
26 tháng 6 2017
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
=>\(2A=-2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
=>\(2A-A=\left(-2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)-\left(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
=>\(A=-2+\frac{1}{1024}\)
NG
1
2 tháng 2 2022
Đặt \(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
\(\Leftrightarrow-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
\(\Leftrightarrow-\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{11}}\)
\(\Leftrightarrow A\cdot\dfrac{1}{2}=\dfrac{1}{2^{11}}-1\)
hay \(A=\dfrac{2\cdot\left(1-2^{11}\right)}{2^{11}}=\dfrac{1-2^{11}}{2^{10}}\)
SH
0
NM
1
TP
28 tháng 7 2017
Ta có:
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-1+\left(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(=-1+\left(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\right)\)
\(=-1+\frac{-1023}{1024}\)
\(=-\frac{2047}{1024}\)