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BT
4
9 tháng 4 2016
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{1024}\right)\)
\(=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{1023}{1024}\)
Còn lại tự làm đi chỉ quy đồng là xong.
U
28 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1028}\)
\(=1-\frac{1}{1028}\)
\(=\frac{1027}{1028}\)
28 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
\(A=\frac{2^{10}-1}{2^{10}}\)
Tham khảo nhé~
NH
4 tháng 7 2017
\(A=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+.......................+\left(1-\dfrac{1}{1024}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+..................+\dfrac{1023}{1024}\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{7}{2^3}+...............+\dfrac{1023}{2^{10}}\)
\(\Leftrightarrow2A=1+\dfrac{3}{2}+\dfrac{7}{2^2}+...............+\dfrac{1023}{2^9}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{3}{2}+\dfrac{7}{2^2}+............+\dfrac{1023}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{7}{2^3}+...............+\dfrac{1023}{2^{10}}\right)\)\(\Leftrightarrow A=1+\left(\dfrac{3}{2}-\dfrac{1}{2}\right)+\left(\dfrac{7}{2^2}-\dfrac{3}{2^2}\right)+.................+\left(\dfrac{1023}{2^9}-\dfrac{511}{2^9}\right)-\dfrac{1023}{2^{10}}\)\(\Leftrightarrow A=\left(1+1+...+1\right)-\dfrac{1023}{2^{10}}\)
\(\Leftrightarrow A=9-\dfrac{1023}{2^{10}}\)
23 tháng 7 2015
gọi A=1/2+1/4+1/8+...+1/1024
2xA=1+1/2+1/4+.....+1/512
2xA-A=(1+1/2+1/4+....+1/512)-(1/2+1/4+1/8+...+1/1024)
A=1-1/1024
=1023/1024
vậy A=1023/1024
\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\)
\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{512}\)
\(2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....+\dfrac{1}{1024}\right)\)
\(A=2-\dfrac{1}{1024}\)
\(A=\dfrac{2047}{1024}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\\ =1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\\ \Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{11}}\\ \Rightarrow A-\dfrac{1}{2}A=1-\dfrac{1}{2^{11}}\\ \Rightarrow A=2-\dfrac{1}{2^{10}}\)