K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
XO
25 tháng 7 2019
Ta có : \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)
= \(\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}\right)\)
< \(\frac{1}{2^2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-\right).n}\right)\)
= \(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
= \(\frac{1}{4}.\left(1-\frac{1}{n}\right)\)
< \(\frac{1}{4}.1=\frac{1}{4}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}.\left(1-\frac{1}{n}\right)< \frac{1}{4}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\left(đpcm\right)\)
3 tháng 4 2016
1/4^2+1/6^2+1/8^2+....+1/(2n)^2<1/4
CMR : Thì nó bé hơn thì cần gì phải chứng minh nhỉ ?
25 tháng 5 2016
1/42+1/62+1/82+...+1/(2n)2
=1/22.22+1/22.32+1/22.42+...+1/22.n2
=1/22.(1/22+1/32+1/42+...+1/n2)<1/22.(1/1.2+1/2.3+1/3.4+...+1/(n-1).n)
<1/4.(1-1/2+1/2-1/3+1/3-1/4+...+1/n-1-1/n)
<1/4.(1-1/n)<1/4
6 tháng 8 2016
1/42+1/62+1/82+...+1/(2n)2
=1/22.22+1/22.32+1/22.42+...+1/22.n2
=1/22.(1/22+1/32+1/42+...+1/n2)<1/22.(1/1.2+1/2.3+1/3.4+...+1/(n-1).n)
<1/4.(1-1/2+1/2-1/3+1/3-1/4+...+1/n-1-1/n)
<1/4.(1-1/n)<1/4
AH
Akai Haruma
Giáo viên
25 tháng 12 2018
Lời giải:
Ta có:
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)
Mà:
\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)
\(< \frac{1}{6}< \frac{1}{4}(**)\)
Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)
6 tháng 3 2019
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)< \frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{n}\right)\)(đpcm)
6 tháng 3 2019
Ta có:\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{4.4}+\frac{1}{4.9}+\frac{1}{4.16}+...+\frac{1}{4.n^2}\)
\(=\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)\)
\(Xét:\)
\(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};\frac{1}{n.n}< \frac{1}{\left(n-1\right).n}...\)
\(Suyra:\)
\(P=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(\Leftrightarrow P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Leftrightarrow P< 1-\frac{1}{n}< 1\)
\(\Leftrightarrow\frac{1}{4}.P< 1.\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{n^2}\right)< \frac{1}{4}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\left(đpcm\right)\)
Đặt
A= \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+\frac{1}{\left(2.4\right)^2}+...+\frac{1}{\left(2n\right)^2}\)
=\(\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}\)
=> \(A=\frac{1}{2^2}\left(1-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)=\frac{1}{4}-\frac{1}{4.n}< \frac{1}{4}\)