Tính \(M=\frac{1+2+2^2+...+2^{2012}}{2^{2014}-2}\)
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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
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Ta có: \(\frac{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...\frac{1}{2014}+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)+1+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+2015}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=\(\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+1\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2015
![](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)
\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
\(=>2M=\frac{2+2^2+2^3+2^4+...+2^{2013}}{2^{2014}-2}\)
\(=>M=\frac{\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)}{2^{2014}-2}\)
\(=>M=\frac{2^{2013}-1}{2^{2014}-2}\)
https://coccoc.com/search/math#query=%5B(2%5E2013)-1%5D%2F%5B(2%5E2014)-2%5D%5D
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
Tính \(A=1+2+2^2+...+2^{2012}\Rightarrow2.A=2.\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow2.A=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow2.A-A=2+2^2+2^3+...+2^{2013}-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(\Rightarrow A=2+2^2+2^3+...+2^{2013}-1-2-2^2-2^3-...-2^{2012}\)
\(\Rightarrow A=2^{2013}-1\)
vậy \(M=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)