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\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\)
\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{512}\)
\(2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\right)\)
\(A=2-\dfrac{1}{1024}\)
\(A=\dfrac{2047}{1024}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\\ =1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\\ \Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{11}}\\ \Rightarrow A-\dfrac{1}{2}A=1-\dfrac{1}{2^{11}}\\ \Rightarrow A=2-\dfrac{1}{2^{10}}\)
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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1028}\)
\(=1-\frac{1}{1028}\)
\(=\frac{1027}{1028}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
\(A=\frac{2^{10}-1}{2^{10}}\)
Tham khảo nhé~
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