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DT
0
14 tháng 9 2016
\(A=\left(1+2\right).\frac{1}{2}+\left(1+2+3\right).\frac{1}{3}+...+\left(1+2+3+...+2016\right).\frac{1}{2016}\)
\(A=\left(1+2\right).2:2.\frac{1}{2}+\left(1+3\right).3:2.\frac{1}{3}+...+\left(1+2016\right).2016:2.\frac{1}{2016}\)
\(A=3:2+4:2+...+2017:2\)
\(A=3.\frac{1}{2}+4.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(A=\frac{1}{2}.\left(3+4+...+2017\right)\)
\(A=\frac{1}{2}.\left(3+2017\right).2015:2\)
\(A=\frac{1}{2}.2020.2015.\frac{1}{2}\)
\(A=505.2015=1017575\)
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
DT
0
18 tháng 6 2020
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2016}\right)\)
\(=\left(1-\frac{1}{\frac{2.3}{2}}\right)\left(1-\frac{1}{\frac{3.4}{2}}\right)...\left(1-\frac{1}{\frac{2016.2017}{2}}\right)\)
\(=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)...\left(1-\frac{2}{2016.2017}\right)\)
\(=\frac{2.3-2}{2.3}.\frac{3.4-2}{3.4}...\frac{2016.2017-2}{2016.2017}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2015.2018}{2016.2017}\)
\(=\frac{1}{3}.\frac{2018}{2016}=\frac{2018}{6048}\)
LN
1
VT
0