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

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

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

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

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

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

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

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

=> \(A=2A-A=1-\frac{1}{2^{10}}\)

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

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

\(=\frac{1}{2^{10}}\)

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

suy ra \(2A=2-1-\frac{1}{2}-...-\frac{1}{512}\)

suy ra \(2A-A=2-\frac{1}{1024}\)

suy ra \(A=\frac{2047}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)

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

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

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

\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)

#)Giải :

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

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

\(\Rightarrow2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

\(\Rightarrow2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{1}{2^{10}}\right)\)

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

\(\Rightarrow A=-\frac{1023}{2048}\)

\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-.....-\frac{1}{1024}\)

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

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

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

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

Ai mún nhờ mk giải violympic vòng 1 300 đ ko

10 nghìn 1 lần nhé hoặc là xóa nick facebook (20 nghìn 1 lần)

Hoặc Hack facebook (10 nghìn 1 lên 500 )

\(\frac{2047}{1024}\) nhé bạn

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

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

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

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

Xin lỗi bạn Trần thị mai Chi nha mk bấm sai kết quả . Kết quả đúng là : 

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

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

Ta có:

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

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

\(2\left[\left(-1\right)-A\right]=1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)

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

\(\left[\left(-1\right)-A\right]=1-\frac{1}{1024}=\frac{1023}{1024}\)

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

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

Ai mún nhờ mk giải violympic vòng 1 300 đ ko

10 nghìn 1 lần nhé hoặc là xóa nick facebook (20 nghìn 1 lần)

Hoặc Hack facebook (10 nghìn 1 lên 500 )

mình chỉ xóa facebook khi facebook sử dụng email và có mk

Đặt :

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

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

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

\(\Leftrightarrow2A-A=\left(2+1+\frac{1}{2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+....+\frac{1}{2^{10}}\right)\)

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

Gọi dãy số trên là A

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

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

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

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

\(A=\frac{2048}{1024}-\frac{1}{1024}\)

\(A=\frac{2047}{1024}\)

Chuyển trừ thành cộng rùi tính.