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26 tháng 8 2015

Chắc mình phải lấy giấy vệ sinh thắt cổ tự tủ mất

28 tháng 10 2017

\(A=1+2+2^2+2^3+...+2^{10}\)

\(2A=2+2^2+2^3+2^4+...+2^{11}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)

\(A=2^{11}-1< 2^{11}\)

\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)

\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)

\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)

\(B=-2^{11}+2^3+10.2^{11}-2^3\)

\(B=9.2^{11}\)

Ta cần so sánh: \(9.2^{11}\)\(2^{14}\)

Hay \(9\)\(2^3\)

\(9>8=2^3\Leftrightarrow B>2^{14}\)

Bài làm 

Đặt a - b = x ; b - c = y ; c - a = z 

 => x + y + z = 0

 Ta có :

          \(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{x+y+z}{xyz}\right)\)

=>     \(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)( Vì x + y + z = 0 )

Vậy ta có đpcm

1 tháng 8 2019

Đặt \(A=2.2^2+3.2^3+4.2^4+5.2^5+...+n.2^n\)

\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+5.2^6+...+n.2^{n+1}\)

\(\Rightarrow2A-A=2.2^3+3.2^4+4.2^5+5.2^6+...+n.2^{n+1}\)

\(-2.2^2-3.2^3-4.2^4-5.2^5-...-n.2^n\)

\(A=n.2^{n+1}-2^3-\left(2^3+2^4+...+2^n\right)\)

Đặt \(M=\left(2^3+2^4+...+2^n\right)\)

\(\Rightarrow2M=\left(2^4+2^5+...+2^{n+1}\right)\)

\(\Rightarrow M=2^{n+1}-2^3\)

\(\Rightarrow A=n.2^{n+1}-2^3-2^{n+1}+2^3\)

\(\Rightarrow A=\left(n-1\right)2^{n+1}=2^{n+10}\)

\(\Rightarrow\left(n-1\right)=2^9\)

\(\Rightarrow n=513\)

1 tháng 8 2019

Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n=2^{n+10}\)

\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)

\(\Rightarrow2A-A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}-2.2^2-3.2^3-4.2^4-...-n.2^n\)

\(\Leftrightarrow A=-2.2^2+\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+[\left(n-1\right)2^n-n.2^n]+n.2^{n+1}\)

\(\Leftrightarrow A=-2.2^2-2^3-2^4-...-2^n+n.2^{n+1}\)

\(\Leftrightarrow A=-2^3-\left(2^4-2^3\right)-\left(2^5-2^4\right)-...-\left(2^{n+1}-2^n\right)+n.2^{n+1}\)

\(\Leftrightarrow A=-2^3-2^4+2^3-2^5+2^4-...-2^{n+1}+2^n+n.2^{n+1}\)

\(\Leftrightarrow A=-2^{n+1}+n.2^{n+1}\)

\(\Leftrightarrow A=2^{n+1}\left(n-1\right)\)

Mà \(A=2^{n+10}=2^{n+1}.2^9=2^{n+1}.512\)

\(\Rightarrow n-1=512\)

\(\Rightarrow n=513\)