A=(1–1/1+2).(1–1/1+2+3)...(1–1/1+2+3+...+2016)
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14 tháng 9 2016
\(A=\left(1+2\right).\frac{1}{2}+\left(1+2+3\right).\frac{1}{3}+...+\left(1+2+3+...+2016\right).\frac{1}{2016}\)
\(A=\left(1+2\right).2:2.\frac{1}{2}+\left(1+3\right).3:2.\frac{1}{3}+...+\left(1+2016\right).2016:2.\frac{1}{2016}\)
\(A=3:2+4:2+...+2017:2\)
\(A=3.\frac{1}{2}+4.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(A=\frac{1}{2}.\left(3+4+...+2017\right)\)
\(A=\frac{1}{2}.\left(3+2017\right).2015:2\)
\(A=\frac{1}{2}.2020.2015.\frac{1}{2}\)
\(A=505.2015=1017575\)
NT
1
18 tháng 8 2016
A=\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{2016^2-1}{2016^2}\)
A=\(\frac{\left(2+1\right)\left(2-1\right)}{2^2}.\frac{\left(3+1\right)\left(3-1\right)}{3^2}......\frac{\left(2016+1\right)\left(2016-1\right)}{2016^2}\)
A=\(\frac{3.4......2017}{2.3....2016}.\frac{1.2...2015}{2.3...2016}\)
A=\(\frac{2017}{2}.\frac{1}{2016}\)
A=\(\frac{2017}{2.2106}>\frac{1}{2}\)
Vậy A\(>\frac{1}{2}\)
PL
1
9 tháng 5 2019
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NX
0
DT
0
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2016}\right)\)
\(=\left(1-\frac{1}{\frac{2.3}{2}}\right)\left(1-\frac{1}{\frac{3.4}{2}}\right)...\left(1-\frac{1}{\frac{2016.2017}{2}}\right)\)
\(=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)...\left(1-\frac{2}{2016.2017}\right)\)
\(=\frac{2.3-2}{2.3}.\frac{3.4-2}{3.4}...\frac{2016.2017-2}{2016.2017}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2015.2018}{2016.2017}\)
\(=\frac{1}{3}.\frac{2018}{2016}=\frac{2018}{6048}\)