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Đặt \(A=1+2+2^2+2^3+...+2^{2012}\)
Có \(2A=2+2^2+2^3+2^4+...+2^{2013}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(A=2^{2013}-1\)
Vậy \(M=\frac{2^{2013}-1}{2.\left(2^{2012}-1\right)}=\frac{1}{2}\)
\(M=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt tử số là A = 1 + 2 + 22 + 23 + ... + 22012
2A = 2 + 22 + 23 + 24 + ... + 22013
2A - A = (2 + 22 + 23 + 24 + ... + 22013) - (1 + 2 + 22 + 23 + ... + 22012)
A = 22013 - 1
=> \(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1 + 2 + 22 + 23+ ...+ 22012
2A = 2 + 22 + 23 + 24 +....+22013
Lấy 2A - A = 2 + 22 +23 + 24 +....+22013 - 1-2-22- 23 - ... - 22012
A = 22013 - 1
Khi đó : M = A / 22014 -2
= 22013 - 1 / 2.( 22013 - 1 )
= 1/2
Vậy M= 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2M=\frac{2+2^2+2^3+...+2^{2013}}{2^{2014}-2}\)
\(2M-M=\frac{\left(2+2^2+...+2^{2013}\right)-\left(1+2^2+...+2^{2012}\right)}{2^{2014}-2}\)
\(M=\frac{1-2^{2013}}{2^{2014}-2}\)
Đặt A=1+2+22+............+22012
2A=2+22+23+..............+22013
2A-A=22013-1
A=22013-1
=>M=\(\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2013}.2-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)
\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)
\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)
\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)