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Khách

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17 tháng 9 2016

ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)

             \(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)

\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)

\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)

\(=1-\frac{1}{1024}\)

\(=\frac{1023}{1024}\)

7 tháng 6 2018

đặt D=-(1/2+1/4+1/8+....+1/1024)

D=-(1/2+1/2^2+....+1/2^10)

đặt A=1/2+...+1/2^10

2A = 1+1/2+...+1/2^9

2A-A=(1+1/2+...+1/2^9)-(1/2+...+1/2^10)

A=1-1/2^10

A=2^10-1/2^10

D=-2^10-1/2^10

2 tháng 2 2020

Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)

Lấy (2) trừ (1) theo vế ta có : 

2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)

13 tháng 9 2016

\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)

\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)

\(A=\frac{1}{1024}\)

13 tháng 9 2016

\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)

\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)

\(B=-\left(1-\frac{1}{1024}\right)\)

\(B=-\frac{1023}{1024}\)

19 tháng 9 2017

Ta có: 

\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

   \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)

\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)

19 tháng 9 2017

\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

Đặt  \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)

Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)

15 tháng 5 2015

Ta có: \(\frac{1}{2}=1-\frac{1}{2}\);  \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\);  ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\)\(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)

Vậy \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

            \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)

            \(=1+1-\frac{1}{1024}\)

            \(=2-\frac{1}{1024}=\frac{2047}{1024}\)

28 tháng 8 2017

bằng 2047/1024

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= \(\frac{1025}{1024}\)

30 tháng 4 2017

= 2047/2048 nha bạn k mình nha 

30 tháng 4 2017

1024 phan 2048

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

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)

\(\Rightarrow A=\frac{513}{1024}\)

10 tháng 5 2017

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)

BẤM ĐÚNG NHÉ

8 tháng 6 2017

1023/1024 nhé bạn