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\(A=1+2+2^2+2^3+...+2^{10}\)
\(2A=2+2^2+2^3+2^4+...+2^{11}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)
\(A=2^{11}-1< 2^{11}\)
\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)
\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)
\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)
\(B=-2^{11}+2^3+10.2^{11}-2^3\)
\(B=9.2^{11}\)
Ta cần so sánh: \(9.2^{11}\) và \(2^{14}\)
Hay \(9\) và \(2^3\)
\(9>8=2^3\Leftrightarrow B>2^{14}\)
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
Đặt \(A=2.2^2+3.2^3+...+n.2^n\)
\(\Rightarrow2A=2.2^3+3.2^4+...+n.2^{n+1}\)
\(\Rightarrow A-2A=2.2^2+\left(3.2^3-2.2^3\right)+...+\left[n.2^n-\left(n-1\right).2^n\right]-n.2^{n-1}\)
\(\Rightarrow-A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)
\(\Rightarrow-A=2+2^1+2^2+2^3+...+2^n-n.2^{n+1}\)
\(\Rightarrow-2A=4+2^2+2^3+...+2^{n+1}-n.2^{n+2}\)
\(\Rightarrow-A-\left(-2A\right)=2+2^1-4-n.2^{n+1}-2^{n+1}+n.2^{n+2}\)
\(\Rightarrow A=n.2^{n+2}-\left(n+1\right)2^{n+1}\)
\(\Rightarrow A=2n.2^{n+1}-\left(n+1\right)2^{n+1}\)
\(\Rightarrow A=\left(n-1\right).2^{n+1}\)
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 \(M=\left(2^3+2^4+...+2^n\right)\)
\(\Rightarrow2M=\left(2^4+2^5+...+2^{n+1}\right)\)
\(\Rightarrow M=2^{n+1}-2^3\)
\(\Rightarrow A=n.2^{n+1}-2^3-2^{n+1}+2^3\)
\(\Rightarrow A=\left(n-1\right)2^{n+1}=2^{n+10}\)
\(\Rightarrow\left(n-1\right)=2^9\)
\(\Rightarrow n=513\)
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 \(A=2^{n+10}=2^{n+1}.2^9=2^{n+1}.512\)
\(\Rightarrow n-1=512\)
\(\Rightarrow n=513\)
\(12^n:2^{2n}=3^n.\left(2^2\right)^n:2^{2n}=3^n.2^{2n}:2^{2n}=3^n\)
\(3^8:3^4+2^2.2^3=3^4+2^5=81+32=113\)
\(\left(7^{1997}-7^{1995}\right)\left(7^{1994}\cdot7\right)=7^{1995}\left(7^2-1\right)\cdot7^{1995}=7^{1995\cdot2}\cdot48=7^{3990}\cdot48\)
\(4^{14}\cdot5^{28}=4^{14}\cdot\left(5^2\right)^{14}=\left(4\cdot25\right)^{14}=100^{14}\)
\(3\cdot4^2-2\cdot3^2=3\cdot2^4-2\cdot3^2=6\left(2^3-3\right)=6\cdot5=30\)
\(18^3:9^3=\left(18:9\right)^3=2^3=8\)
\(\left(2^8+8^3\right):\left(2^5\cdot2^3\right)=\left(2^8+2^9\right):2^8=\dfrac{2^8}{2^8}+\dfrac{2^9}{2^8}=1+2=3\)
\(16\cdot64\cdot8^2:\left(4^3.2^5.16\right)=2^4\cdot2^6\cdot2^6:\left(2^6\cdot2^5\cdot2^4\right)=2\)
\(5\cdot2^9\cdot6^{10}-7\cdot2^{29}\cdot27^6=5\cdot2^9\cdot2^{10}\cdot3^{10}-7\cdot2^{29}\cdot3^{18}=2^{19}\cdot3^{10}\left(5\cdot3^{10}-7\cdot2^{10}\cdot8\right)\)