\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\right)\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(B=1-\frac{1}{2^{2012}}\)

\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{2012}}\)

đúng rùi đó

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)

\(A=2-\frac{1}{2^{2012}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

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 \(A=2-\frac{1}{2^{2012}}\)

\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)

=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)

=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)

=>\(A=2-\frac{1}{2^{2012}}\)

Cô mình chữa bài này rồi nên bạn cứ yên tâm

Xét mẫu số ta có: \(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\) 

=\(2012+\left(\frac{2014-2}{2}+\frac{2014-3}{3}+...+\frac{2014-2013}{2013}\right)\) 

= \(2012+\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}\right)-\left(\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2013}{2013}\right)\) 

= \(2012+2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2012\) 

= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\) 

\(\Rightarrow A=\frac{1}{2014}\)

\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)

\(2A=2+1+...+\frac{2}{2^{2011}}\)

\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

Ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)

=>  \(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)

=>  \(2A=2+1+...+\frac{2}{2^{2011}}\)

=> \(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)

=> \(A=2-\frac{1}{2012}\)

A= 1+ 1/2 + 1/2² + ... + 1/2²⁰¹² 

﴾1/2﴿A= 1/2+1/2²+...+1/2²⁰¹³

A‐﴾1/2﴿A= ﴾1+ 1/2 + 1/2² + ... + 1/2²⁰¹² ﴿ ‐ ﴾ 1/2+1/2²+...+1/2²⁰¹³ ﴿

﴾1/2﴿A = 1 ‐ 1/2²⁰¹³ 

A= ﴾1‐ 1/2²⁰¹³ ﴿ : 1/2

A= 2 ‐ 1/2²⁰¹²

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     = \(2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

mk nhanh nhat nhe

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

\(=2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{2011}}-\frac{1}{2^{2012}}=2-\frac{1}{2^{2012}}\)