1. Chứng minh:
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{\left(2n\right)^2}<\frac{1}{2}\)
2. Tìm 2 số nguyên x, y biết:
36 - y2 = 11(x - 2014)2
3. Tính: \(\frac{423134.846267-423133}{423133.846267-423134}\)
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Những câu hỏi liên quan
HT
12 tháng 3 2017
a. Ta có: \(\frac{1}{2^2}\)< \(\frac{1}{1.3}\)
\(\frac{1}{4^2}\)< 1/(3.5)
1/(6^2) <1/(5.7)
...
1/(2n)^2 < 1/(2n-1)(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 1/(1.3) +...+1/(2n-1)(2n+1)
=> 2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < (1/1 - 1/3 +1/3 - 1/5 + 1/5 - 1/7 +...+ 1/(2n-1) - 1/(2n+1)
=>2(1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2) < 1 - 1/(2n+1) = 2n/(2n+1)
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 < 2n/(2n+1) . 1/2
Vì 2n/2n+1 < 1 => 2n/(2n+1) . 1/2 < 1/2
=> 1/2^2 +1/4^2 + 1/6^2 +...+1/(2n)^2 <1/2
Câu b tương tự
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 9
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
< \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
< \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
< \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
< \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
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 = \(\frac{1}{4}\)
\(\Rightarrow\)\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{4}\)
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 \(A< \frac{1}{4}\)
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:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\)
\(\Rightarrow A< \frac{1}{2^2}.1=\frac{1}{4}\)