Đáp án: $\dfrac BA=2017$
Giải thích các bước giải:
Ta có:
$B=\dfrac{2016}{1}+\dfrac{2015}{2}+...+\dfrac{2}{2015}+\dfrac{1}{2016}$
$\to B=2016+\dfrac{2015}{2}+...+\dfrac{2}{2015}+\dfrac{1}{2016}$
$\to B=1+(1+\dfrac{2015}{2})+...+(1+\dfrac{2}{2015})+(1+\dfrac{1}{2016})$
$\to B=1+\dfrac{2+2015}{2}+...+\dfrac{2015+2}{2015}+\dfrac{2016+1}{2016}$
$\to B=1+\dfrac{2017}{2}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}$
$\to B=\dfrac{2017}{2}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}+1$
$\to B=\dfrac{2017}{2}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}+\dfrac{2017}{2017}$
$\to B=2017(\dfrac12+...+\dfrac1{2015}+\dfrac1{2016}+\dfrac1{2017})$
$\to B=2017A$
$\to\dfrac BA=2017$
