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19 tháng 3 2017

bn đâu có phải hotgirl đâu

19 tháng 3 2017

Ta có:

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)

\(\frac{1}{8^2}=\frac{1}{8.8}< \frac{1}{7.8}\)

\(...\)

\(\frac{1}{\left(2n\right)^2}=\frac{1}{2n.2n}< \frac{1}{1n.2n}\)

Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{1n.2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1n}-\frac{1}{2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}+\left(\frac{-1}{4}+\frac{1}{4}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+...-\frac{1}{2n}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(< \)\(\frac{1}{3}-\frac{1}{2n}\)

6 tháng 5 2018

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\) ta có : 

\(A=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}\)

\(A=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+\frac{1}{2^2}.\frac{1}{4^2}+...+\frac{1}{2^2}.\frac{1}{n^2}\)

\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

\(A< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)

\(A< \frac{1}{2^2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(A< \frac{1}{2^2}\left(1-\frac{1}{n}\right)< \frac{1}{2^2}.1\)

\(A< \frac{1}{2^2}=\frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

Chúc bạn học tốt ~ 

6 tháng 5 2018

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)

\(=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{\left(2n-2\right)\cdot2n}\)

\(=\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{\left(2n-2\right)\cdot2n}\right)\cdot\frac{1}{2}\)

\(=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\cdot\frac{1}{2}\)

\(=\left(\frac{1}{2}-\frac{1}{2n}\right)\cdot\frac{1}{2}=\frac{1}{4}-\frac{1}{2n\cdot2}< 1\)

\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)

18 tháng 8 2017

\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\).... \(+\frac{1}{\left(2n\right)^2}\)\(\frac{1}{2^2}\). ( \(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}\)) < \(\frac{1}{2^2}\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).\left(n\right)}\)) = \(\frac{1}{2^2}\)\(1-\frac{1}{n}\)) < \(\frac{1}{2^2}\).1 = \(\frac{1}{4}\)

\(\Rightarrow\)\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)\(\frac{1}{4}\)

3 tháng 4 2020

mình ko hiểu lắm

8 tháng 7 2016

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

Có:

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(...\)

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)

\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)

8 tháng 4 2017

bài này tui bít làm nhưng dài lắm

8 tháng 4 2017

Cố gắng làm hộ mình với !

26 tháng 2 2018

Ta có : 

\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)

\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)

\(A< \frac{1}{4}-\frac{1}{4n}\)

Lại có \(n>0\) nên \(\frac{1}{4n}>0\)

\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

\(\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{n^2}

2 tháng 4 2017

trên là 2n ở dưới lại là n