cách này ngon hơn nè
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
=> 2A = 1/2^1+1/2^2+...+1/2^2012+1/2^2013
=> 2A - A = 1A =(1/2^1+1/2^2+...+1/2^2012+1/2^2013)-(1/2^0+1/2^1+1/2^2+1/2^3+...+1/2^2012)
=> 1A = 1/2^2012-1/2^0
= 1/2^2012-1/1
Đúng rùi đó cô mk cũng chữa như z