CMR:
\(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+..................+\frac{1}{64^2}< \frac{5}{16}\)
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T
30 tháng 4 2019
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
ML
19 tháng 3 2017
P = 1/4 + 1/16 + 1/36 + .. + 1/196 = 1/2² + 1/4² + 1/6² +...+ 1/12² + 1/14²
xét tổng quát với số nguyên dương k ta có:
(2k-1)(2k+1) = 4k² - 1 < 4k² = (2k)² => 1/(2k)² < 1/(2k-1)(2k+1)
=> 2/(2k)² < 2 /(2k-1)(2k+1) = 1/(2k-1) - 1/(2k+1) (*)
ad (*) cho k từ 1 đến 7
2/2² < 1/1 - 1/3
2/4² < 1/3 - 1/5
...
2/12² < 1/11 - 1/13
2/14² < 1/13 - 1/15
+ + cộng lại + +
2/2² + 2/4² +...+ 2/14² < 1/1 - 1/15 < 1
=> 2(1/2² + 1/4² +..+ 1/14²) < 1 => P < 1/2 (đpcm)
DL
25 tháng 6 2018
b) \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{13\cdot15}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(1-\frac{1}{15}\right)< \frac{1}{2}\)
\(\)
A
25 tháng 6 2018
a, \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}< 1\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(........\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{n}\)\(=1-\frac{1}{n}=\frac{n-1}{n}< 1\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)
Đặt \(B=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
....................
\(\frac{1}{64^2}< \frac{1}{63.64}\)
\(\Rightarrow B< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)
\(\Rightarrow B< \frac{1}{4}-\frac{1}{64}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{4^2}+\frac{1}{4}\)
\(\Rightarrow A< \frac{5}{16}\)
Ta có S =\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)
= \(\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{64.64}\)
< \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)
= \(\frac{1}{3}-\frac{1}{64}\)
= \(\frac{61}{192}\)> \(\frac{60}{192}=\frac{5}{16}\)
S < \(\frac{61}{192}>\frac{5}{16}\)
=> sai đề