Ta có :

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

    \(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)

    \(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)

    \(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\) 

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)

    \(=2015.2.\left(1-\frac{1}{2017}\right)\)

    \(=2015.2.\frac{2016}{2017}\)

    =\(\frac{2015.2.2016}{2017}\)

    =\(\frac{8124480}{2017}\)

Vậy \(S=\frac{8124480}{2017}\)

Sai vì ngoài học tập ra còn cần phải siêng năng chăm chỉ trong các lĩnh vực khác nửa như giúp đỡ mọi người ,tham gia các hoạt động rèn luyện

a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)

b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)

bằng 15 hay sao ý

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

\(\Rightarrow3B=3+3^2+3^3+3^4+...+3^{2016}-\frac{3^{2017}}{2}\)

\(\Rightarrow2B=3^{2016}-\frac{3^{2017}}{2}-1+\frac{3^{2016}}{2}\)

\(=3^{2016}-1-\left(\frac{3^{2017}}{2}-\frac{3^{2016}}{2}\right)\)

\(=3^{2016}-1-\frac{3^{2017}-3^{2016}}{2}\)

\(\Rightarrow B=\frac{\left(3^{2016}-1-\frac{3^{2017}-3^{2016}}{2}\right)}{2}\)

<br class="Apple-interchange-newline"><div></div>B=1+3+3²+3³+3⁴+...+3²⁰¹⁵−3²⁰¹⁶2 

Đặt \(A=1+3+3^2+...+3^{2015}\)

\(A=1+3+3^2+...+3^{2015}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2016}\)

\(\Rightarrow2A=3^{2016}-1\)

\(\Rightarrow A=\frac{3^{2016}-1}{2}\)

\(\Rightarrow B=\frac{3^{2016}-1-3^{2016}}{2}=\frac{-1}{2}\)

lỗi hiển thị 1 chút,bn thông cảm

wow lấy ở đâu zậy
 

Đăng từng bài thôi         

=2015/2016

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

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)

\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)

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