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\(\Rightarrow\frac{M}{512}=1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\)
\(\Rightarrow2.\left(\frac{M}{512}\right)=2-1-\frac{1}{2}-.....-\frac{1}{2^9}\)
\(\Rightarrow2.\left(\frac{M}{512}\right)-\frac{M}{512}=\left(2-1-\frac{1}{2}-.....-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\right)\)
\(\Rightarrow\frac{M}{512}=-\frac{1}{2^{10}}\)
\(\Rightarrow M=-\frac{1}{2}\)
M= 512 - \(\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
=> 2.M = 1024 - 512 - \(\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^9}\)
=> 2.M - M = 1024 - 512 - 512 + \(\frac{512}{2^{10}}\)
=> M = \(\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)
M = \(512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-.....-\frac{512}{2^{10}}\)
M = \(512-512.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
2A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{11}}\)
A = 2A - A = \(1-\frac{1}{2^{10}}\)
=> M = \(512-512.\left(1-\frac{1}{2^{10}}\right)\)
=> M = 512.\(\left(1-1+\frac{1}{2^{10}}\right)\)
=> M = \(512.\frac{1}{2^{10}}\)
=> M = \(\frac{512}{2^{10}}\)
Ta có:
\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(\Rightarrow P=512-\left(\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\right)\)
Đặt \(A=\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\)
\(\Rightarrow2A=512+\frac{512}{2}+\frac{512}{2^2}+...+\frac{512}{2^9}\)
\(\Rightarrow2A-A=512-\frac{512}{2^{10}}\)
\(\Rightarrow A=512-\frac{512}{2^{10}}\)
\(\Rightarrow P=512-A=512-\left(512-\frac{512}{2^{10}}\right)=\frac{512}{2^{10}}=\frac{1}{2}\)
\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(\Rightarrow P=2^9-\frac{2^9}{2}-\frac{2^9}{2^2}-\frac{2^9}{2^3}-...-\frac{2^9}{2^{10}}\)
\(\Rightarrow P=2^9-2^8-2^7-2^6-...-\frac{1}{2}\)
\(\Rightarrow2P=2^{10}-2^9-2^8-2^7-...-1\)
\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-\left(2^9-2^8-2^7-2^6-...-\frac{1}{2}\right)\)
\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-2^9+2^8+2^7+2^6+...+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2^9-2^9+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2.2^9+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2^{10}+\frac{1}{2}\)
\(\Rightarrow P=0+\frac{1}{2}\)
\(\Rightarrow P=\frac{1}{2}.\)
Chúc bạn học tốt!
\(B=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\\ =512\cdot\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\right)\\ =512\cdot\left[1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\right]\)
Đặt \(H=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\Leftrightarrow B=512\cdot\left(1-H\right)\)
\(\Leftrightarrow2H=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\\ \Leftrightarrow2H-H=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\\ \Leftrightarrow H=1-\frac{1}{2^{10}}\\ \Leftrightarrow B=512\cdot\left[1-\left(1-\frac{1}{2^{10}}\right)\right]\\ \Leftrightarrow B=512\cdot\frac{1}{2^{10}}\\ \Rightarrow B=2^9\cdot\frac{1}{2^{10}}\\ \Rightarrow B=\frac{1}{2}\)
M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2
M = 1/2
M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2
M = 1/2