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Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)
\(=\frac{2015}{2}+1+\frac{2014}{3}+1+...........+\frac{1}{2015}+1\)
\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)
\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)
Thay A vào biểu thức ta dc
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)
\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)
CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=\frac{\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=\frac{\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}}\)
\(B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=2016\)
\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{1+\left(\frac{2014}{2}+1\right)+\left(\frac{2013}{3}+1\right)+\left(\frac{2012}{4}+1\right)+...+\left(\frac{1}{2015}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{\frac{2016}{2016}+\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=2016\)
Vậy \(B=2016\)