bằng 15 hay sao ý

xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)

=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)

=2016/2015+2016/2014+...+2016/3+2016/2+1

=2016.(1/2016+1/2015+...+1/4+1/3+1/2)

=> A= 1/2016

mún dễ hỉu hơn hãy gửi tin nhắn cho mik

1 phan 2016. cac lam de lam

mà +1 ở phần tử hay mẫu z

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

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 \(B=2016\)

Có \(B=\frac{\frac{2015}{1}+\frac{2014}{2}+......+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2016}}\)

Xét mẫu số:

Đặt A là mẫu số; C là tử số

\(A=\frac{2015}{1}+\frac{2014}{2}+......+\frac{1}{2015}\)

\(=\left(\frac{1}{2015}+1\right)+\left(\frac{2}{2014}+1\right)+.........+\left(\frac{2015}{1}-2014\right)\)

\(=\frac{2016}{2015}+\frac{2016}{2014}+.........+\frac{2016}{2016}\)

\(=2016.\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+..........+\frac{1}{2}\right)\)

\(=2016.C\)

\(\Rightarrow B=\frac{2016.C}{C}=2016\)

Vậy B = 2016