\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
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NH
1
TN
25 tháng 5 2016
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
WH
8 tháng 4 2018
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(B=1-\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
T
2
11 tháng 5 2019
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)
\(A=2-\frac{1}{2^{2012}}\)
13 tháng 4 2017
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
Nên \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Suy ra \(2A-A=2-\frac{1}{2^{2012}}\)hay \(A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
13 tháng 4 2017
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)
=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
=>\(A=2-\frac{1}{2^{2012}}\)
Cô mình chữa bài này rồi nên bạn cứ yên tâm
TN
17 tháng 5 2016
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+...+\frac{2}{2^{2011}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
17 tháng 5 2016
Ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
=> \(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(2A=2+1+...+\frac{2}{2^{2011}}\)
=> \(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2012}\)
8 tháng 4 2016
Xét mẫu số ta có: \(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
=\(2012+\left(\frac{2014-2}{2}+\frac{2014-3}{3}+...+\frac{2014-2013}{2013}\right)\)
= \(2012+\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}\right)-\left(\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2013}{2013}\right)\)
= \(2012+2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2012\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow A=\frac{1}{2014}\)
HL
1
TL
1
12 tháng 8 2015
Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
2A = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
2A - A = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-..-\frac{1}{2^{2011}}-\frac{1}{2^{2012}}\)
A = \(2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)