Cho :X=2+2^2+2^3+......................+2^2013 và Y =2^2014
so sánh X Và Y
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\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)
\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)
\(A=\frac{-2015}{4028}\)
\(y=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2014^2}-1\right)\)
\(y=\left(\frac{-1.3}{2.2}\right)\left(\frac{-2.4}{3.3}\right)....\left(\frac{-2013.2015}{2014.2014}\right)\)
\(y=-\left(\frac{1.2....2013.3.4...2015}{2.3....2014.2.3....2014}\right)\)
\(y=-\left(\frac{2015}{2014.2}\right)\)
\(y=\frac{-2015}{4028}\)
\(x=\frac{-1}{2}=\frac{-2014}{4028}\)
Vì \(\frac{-2015}{4028}
Ta có: \(A=\dfrac{x-y}{x+y}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\)
\(=\dfrac{x^2-y^2}{x^2+2xy+y^2}\)
Ta có: \(x^2+2xy+y^2>x^2+y^2\forall x>y>0\)
\(\Leftrightarrow\dfrac{x^2-y^2}{x^2+2xy+y^2}< \dfrac{x^2-y^2}{x^2+y^2}\)
hay A<B
Ta có X=2+22+23+...+22013
=>2X=2*(2+22+23+...+22013)
=>2X=22+23+24+...+22014
=>X-2X=2+22+23+...+22013-(22+23+24+...+22014)
=>X=2+22+23+...+22013-22-23-24-...-22014
=>X=2-22014
Ta có : 2-22014 < 22014
=>X<Y
Vậy X<Y
Tích mk nha .....