K
Khách
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KN
1
3 tháng 7 2018
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2014}\)
\(\Rightarrow-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow-2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow-2A-\left(-A\right)=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)
\(-A=2-\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2^{10}}-2\)
T
24 tháng 1 2017
đặt A = \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
ta có:
A = \(-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
A = \(-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
Đặt B = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
ta có:
B = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
=> 2B = \(2+1+\frac{1}{2}+...+\frac{1}{512}\)
=> 2B - B = \(\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> B = \(2-\frac{1}{1024}\)
=> B = \(\frac{2048}{1024}-\frac{1}{1024}=\frac{2047}{1024}\)
Thay B vào A ta có:
A = \(\frac{-2047}{1024}\)
vậy A = \(\frac{-2047}{1024}\)
SN
0
AM
1
NT
1
HT
22 tháng 2 2017
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)
=> \(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2B=2+1+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(B=2B-B=2-\frac{1}{2^{10}}\)
=> \(A=-\left(2-\frac{1}{2^{10}}\right)=-\left(2-\frac{1}{1024}\right)=-\frac{2047}{1024}\)
17 tháng 7 2019
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
17 tháng 7 2019
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
NN
28 tháng 12 2016
-1-1/2-1/4-1/8......-1/1024
=-(1+1/2+1/4+1/8...+1/1024)
mà ta có 1024=2^10
nên -(1+1/2+1/4+1/8...+1/1024)
=-(2^9+2^8+2^7....+1)/2^10
=-(1023/1024)
=-1,99.........
NN
19 tháng 1 2017
mình sẽ làm lại bai này cho đúng nha
\(-1-\frac{1}{2}-\frac{1}{4}....-\frac{1}{1024}=-1-\left(\frac{1}{2}+\frac{1}{4}+...\frac{1}{1024}\right)\)
\(=-1-\left(\frac{1}{2^1}+\frac{1}{2^2}...+\frac{1}{2^{10}}\right)\)
\(=-1-\frac{1023}{1024}=\frac{-1024}{1024}-\frac{1023}{1024}=\frac{-2047}{1024}\)
vậy mới đúng nha
HT
0
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}\)
Đặt:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)
\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}\)
\(A=1-\dfrac{1}{1024}=\dfrac{2023}{2024}\)