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Bản đẹp :
CMR : \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+.....+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
N = \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2n\right)^2}\)
= \(\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\) (lát nữa mình sẽ chứng minh)
=> N <\(\frac{1}{4}.1=\frac{1}{4}\)
Ta sẽ chứng minh bổ đề: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\)
Thật vậy:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)= \(1-\frac{1}{n^2}<1\)
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A=1/4^2+1/6^2+...+1/(2n)^2
=1/4(1/2^2+1/3^2+...+1/n^2)
=>A<1/4(1-1/2+1/2-1/3+...+1/n-1-1/n)
=>A<1/4(1-1/n)<1/4
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Ta có:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)\)
\(=\frac{1}{4}-\frac{1}{2n.2}\)
Vì \(\frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\) (Đpcm)
cậu biến đổi 1/4^2 thành 1/(2.2)^2; 1/6^2 thành 1/(2.3)^2 ấy để cho nó có dạng 1/(2n)^2 là sẽ giải ra thôi
chứ tớ ngại ghi ra vì dài lắm
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