Cho A=(1/2^2-1)x(1/3^2-1)x(1/4^2-1)...(1/2013^2-1)x(1/2014^2-1) và B=-1/2.so sánh A và B
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QH
Cho A = (1/2^2 - 1)(1/3^2 - 1) (1/4^2 - 1) ... (1/2013^2 -1)(1/2014^2 - 1) Và B = -1/2
So sánh A và B
0
NH
2
16 tháng 8 2015
$A=\frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{2014^2-1}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}=1-\frac{1}{2014}=\frac{2013}{2014}>-\frac{1}{2}$
BB
0
17 tháng 8 2021
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
QK
0
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)
\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)
\(A=\frac{-2015}{4028}\)