1.A=1/2 + 1/3 + 1/4 + … + 1/2016 + 1/2017 B=2016/1 + 2015/2 +…+2/2015 +1/2016 Tính B/A 14/09/2021 Bởi Faith 1.A=1/2 + 1/3 + 1/4 + … + 1/2016 + 1/2017 B=2016/1 + 2015/2 +…+2/2015 +1/2016 Tính B/A
Đáp án: $\dfrac BA=2017$ Giải thích các bước giải: Ta có: $B=\dfrac{2016}{1}+\dfrac{2015}{2}+…+\dfrac{2}{2015}+\dfrac{1}{2016}$ $\to B=2016+\dfrac{2015}{2}+…+\dfrac{2}{2015}+\dfrac{1}{2016}$ $\to B=1+(1+\dfrac{2015}{2})+…+(1+\dfrac{2}{2015})+(1+\dfrac{1}{2016})$ $\to B=1+\dfrac{2+2015}{2}+…+\dfrac{2015+2}{2015}+\dfrac{2016+1}{2016}$ $\to B=1+\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}$ $\to B=\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}+1$ $\to B=\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}+\dfrac{2017}{2017}$ $\to B=2017(\dfrac12+…+\dfrac1{2015}+\dfrac1{2016}+\dfrac1{2017})$ $\to B=2017A$ $\to\dfrac BA=2017$ Bình luận
Đáp án: $\dfrac BA=2017$
Giải thích các bước giải:
Ta có:
$B=\dfrac{2016}{1}+\dfrac{2015}{2}+…+\dfrac{2}{2015}+\dfrac{1}{2016}$
$\to B=2016+\dfrac{2015}{2}+…+\dfrac{2}{2015}+\dfrac{1}{2016}$
$\to B=1+(1+\dfrac{2015}{2})+…+(1+\dfrac{2}{2015})+(1+\dfrac{1}{2016})$
$\to B=1+\dfrac{2+2015}{2}+…+\dfrac{2015+2}{2015}+\dfrac{2016+1}{2016}$
$\to B=1+\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}$
$\to B=\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}+1$
$\to B=\dfrac{2017}{2}+…+\dfrac{2017}{2015}+\dfrac{2017}{2016}+\dfrac{2017}{2017}$
$\to B=2017(\dfrac12+…+\dfrac1{2015}+\dfrac1{2016}+\dfrac1{2017})$
$\to B=2017A$
$\to\dfrac BA=2017$