\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\)

<\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)

Ta có:\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}\)

=1

=>\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\) \(

Ta có : \(\frac{1}{2^2}

Katherine làm sai cmnr \(\frac{1}{2^2}\)giải kiểu gì ra\(\frac{1}{1.2}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

                                                                            \(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

\(< 1-\frac{1}{2016}< 1\left(đpcm\right)\)

Thằng vua hải tặc vàng oai vừa thôi !

thi sao

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

\(\Leftrightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2014.2025}\)

\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2014.2015}\)

\(\Leftrightarrow B< 1-\frac{1}{2015}< 1\)

\(\Leftrightarrow B< 1\rightarrowđpcm\)

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)

+ Xét : \(\frac{1}{1\cdot2}>\frac{1}{2^2}\)

\(\frac{1}{2\cdot3}>\frac{1}{3^2}\)

\(\frac{1}{3\cdot4}>\frac{1}{4^2}\)

...

\(\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(A=1-\frac{1}{2015}< 1\)

\(\Rightarrow B< A< 1\left(đpcm\right)\)

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(=1-\frac{1}{2016}=\frac{2015}{2016}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)

​\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{2016}-\frac{1}{2017}\)

\(=\frac{1}{2}-\frac{1}{2017}=\frac{2015}{4024}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{2015}{4034}\)

vậy ta có điều cần chứng minh

Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)

\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2014\cdot2015}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}< 1\)

Vậy \(A< 1\left(đpcm\right)\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2015^2}< \frac{1}{2014.2015}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(=1-\frac{1}{2015}< 1^{\left(đpcm\right)}\)

tớ cũng không biết đâu .Nếu tìm ra cách giải thì nhắn tin cho tớ nha

Bài này trước tiên ta phải đi chứng minh công thức:

                      \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

 Xong áp dụng là ra thui.
 

Hay thật Công thức rất hay c/m không phức tạp lắm.

nhưng từ bài toán ban đầu tự nhiên nội suy ra được cái công thức đó. Khó nhỉ