\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

xam xi

Đặt N = 1 + 2 + 2² +...+ 2²⁰¹²

2N = 2 + 2² + 2³ +...+ 2²⁰¹³

2N - N = (2 + 2² + 2³+....+ 2²⁰¹³) - (1 + 2 + 2² +....+ 2²⁰¹²)

N = 2²⁰¹³ - 1

Thay N vào M ta được:

Đặt \(N=1+2+2^2+...+2^{2012}\)

\(2N=2+2^2+2^3+...+2^{2013}\)

\(2N-N=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)

\(N=2^{2013}-1\)

Thay N vào M ta được:

\(M=\dfrac{2^{2013-1}}{2^{2014}-2}=\dfrac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\dfrac{1}{2}\)

Sửa đề: \(M=\frac{1+2+2^2+\cdots+2^{2012}}{2^{2014}-2}\)

Đặt \(A=1+2+2^2+\cdots+2^{2012}\)

=>\(2A=2+2^2+2^3+\cdots+2^{2013}\)

=>2A-A=\(2+2^2+2^3+\cdots+2^{2013}-1-2-2^2-\cdots-2^{2012}\)

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

Ta có: \(M=\frac{1+2+2^2+\cdots+2^{2012}}{2^{2014}-2}\)

\(=\frac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\frac12\)

Đặt \(A=2^{2013}+2^{2012}+\cdots+2^3+2^2+3\)

=>\(A=2^{2013}+2^{2012}+\cdots+2^2+2+1\)

=>\(2A=2^{2014}+2^{2013}+\cdots+2^3+2^2+2\)

=>\(2A-A=2^{2014}+2^{2013}+\cdots+2^3+2^2+2-2^{2013}-2^{2012}-\cdots-2^2-2-1\)

=>\(A=2^{2014}-1\)

Ta có: \(B=2^{2014}-2^{2013}-2^{2012}-\cdots-2^3-2^2-3\)

=>\(B=2^{2014}-\left(2^{2014}-1\right)\)

\(=2^{2014}-2^{2014}+1=1\)

\(2P=2+2^2+2^3+...+2^{2022}\)

\(\Leftrightarrow P=2^{2022}-1< Q\)

\(2P=2+2^2+2^3+...+2^{2022}\)

\(\Leftrightarrow P< Q\)

23/27> 22/29

23/27>22/29

Ta lấy ps trung gian: 23/29

So sánh : 23/27>23/29>22/29.

=> 23/27>22/29