\(M=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(M=512-\frac{512}{2}-\frac{512}{4}-\frac{512}{8}-...-\frac{512}{1024}\)
\(M=\frac{1024}{2}-\frac{512}{2}-\frac{256}{2}-\frac{128}{2}-...-\frac{1}{2}\)
\(M=\frac{1024}{2}-\left(\frac{512}{2}+\frac{256}{2}+\frac{128}{2}+\frac{64}{2}+...+\frac{1}{2}\right)\)
\(M=\frac{1024}{2}-\frac{1023}{2}\)
\(M=\frac{1}{2}\)
\(M=0,5\)
\(M=512-\frac{512}{2^2}-....-\frac{512}{2^{10}}\)
\(=2^9-\frac{2^9}{2}-.....-\frac{2^9}{10}\)
\(=2^9-2^8-....-\frac{1}{2}\)
\(2M=2^{10}-2^9-....-1\)
\(M=\left(2^{10}-...-1\right)-2^9+2^8+....+1+\frac{1}{2}\)
\(M=2^{10}-2.2^9+\frac{1}{2}\)
\(M=\frac{1}{2}\)
Ta có :
M= \(512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)=\(512\left(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
Đặt A= \(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)(1)
=> 2A=\(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)(2)
Lấy (2)-(1) ta được:
A= \(2+\frac{1}{2^{10}}=\frac{2^{11}+1}{2^{10}}\)
=> M= \(512\cdot\frac{2^{11}+1}{2^{10}}=2^9\cdot\frac{2^{11}+1}{2^{10}}\)= \(\frac{2^{11}+1}{2}=2^{10}+\frac{1}{2}=1024+\frac{1}{2}=\frac{2049}{2}=1024,5\)
Vậy M= \(1024,5\)
Làm lại:
Ta có:
M= \(512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}=512\left(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
Đặt A= \(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)(1)
=> 2A=\(2-1-\frac{1}{2}-...-\frac{1}{2^9}\)(2)
Lấy (2)-(1) ta được:
A=\(\frac{1}{2^{10}}\)
=> M= \(512\cdot\frac{1}{2^{10}}=2^9\cdot\frac{1}{2^{10}}=\frac{1}{2}=0,5\)
Vậy M=0,5