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Đặt A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+................+\frac{1}{2048}+\frac{1}{4096}\)
2A=\(1+\frac{1}{2}+\frac{1}{4}+....................+\frac{1}{1024}+\frac{1}{2048}\)
2A-A=\(\left(1+\frac{1}{2}+\frac{1}{4}+................+\frac{1}{1024}+\frac{1}{2048}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...............+\frac{1}{2048}+\frac{1}{4096}\right)\)
A=\(1-\frac{1}{4096}\)
A=\(\frac{4095}{4096}\)
Từ biểu thức trên ta được
A=1/2+1/2^2+1/2^3+....+1/2^11+1/2^12
2A-A=2(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)-(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)
A=1+1/2^2+1/2^3+....+1/2^11-1/2-1/2^2-...-1/2^12
A=1/2-1/2^12
Ủng hộ cho mình nha bạn
\(A=\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}+\frac{1}{4096}\)
\(A=\)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)
\(2A=\)\(2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(2A=\)\(1+\frac{1}{2}+...+\frac{1}{2^{11}}\)
\(2A-A=\)\(\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(A=1-\frac{1}{2^{12}}\)
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
Đặt tổng trên là A . Ta có:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
Ta co: 2S=2+1+1/2+1/4+...+1/2048
2S-S=2+1+1/2+1/4+...+1/2048-1-1/2-1/4-...-1/2048-1/4096
\(\Rightarrow\)S=2-1/4096 =8191/4096
\(\text{Đặt }S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}.\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(\Rightarrow2S-S=S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Rightarrow S=1-\frac{1}{2048}=\frac{2047}{2048}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}+\frac{1}{4096}\)
\(\Leftrightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}\)
Nhân 2 vào 2 vế của biểu thức A , ta được :
\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}\right)\)
\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^9}+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)
Lấy biểu thức 2A - A , Ta được :
\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}+\frac{1}{2^{12}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{12}}\)