Question

[Vũ Việt Hà]

CMR:

\(\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+...+\frac{1}{2013^2}< \frac{1}{5}\)

Answer 1

[nhok ma kết]

\(A< \frac{1}{1\cdot3}+\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+..........+\frac{1}{2011\cdot2013}\)

\(\frac{1}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+.....+\frac{1}{2010}-\frac{1}{2013}\right)\)

\(\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}\cdot\frac{2012}{2013}\)

theo mình là vậy thôi chứ ko chắc chắn đouo

Answer 2

[Tony Tony Chopper]

bạn nhok ma kết làm gần đúng nhưng vẫn sai nhé

Đặt biểu thức là A

\(A=\frac{1}{9}\left(\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{671^2}\right)< \frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{671.672}\right)\)

\(\Rightarrow A< \frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{671}-\frac{1}{672}\right)\)

\(\Rightarrow A< \frac{1}{9}\left(1-\frac{1}{672}\right)=\frac{1}{9}.\frac{671}{672}< \frac{1}{5}.1=\frac{1}{5}\)