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Ta có A = 3^2015 - 2^2015 + 3^2013 - 2^2013
= 3^2015 + 3^2013 - ( 2^2015 + 2^2013)
= 3^2013.3^2 + 3^2013 - ( 2^2013.2^2 + 2^2013)
= 3^2013.(3^2+1) - 2^2013.(2^2+1)
= 3^2013.10 - 2^2013.5
= 3^2013.2.5 - 2^2013.5
= 5 . (3^2013.2 - 2^2013) chia hết cho 5
Vậy A chia hết cho 5
A= 3+3^2+3^3+.....+3^2015+3^2016
2A=3^2+3^4+........+3^2016 +2^2017
2A-A= (3^2-3^2) + ( 3^3-3^3)+..........+(3^2015-3^2015)+(3^2016-3^2016)+(3^2017 -3)
A= 3 ^2017 - 3
Hết
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NHân với 3