Cho A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2011}{2012}\). Chứng minh: A2 < \(\frac{1}{2013}\).
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GW
12 tháng 9 2021
a )
Theo bài ra: (a - 4) chia hết cho 5 => (a - 4) + 20 chia hết cho 5 => a + 16 chia hết cho 5
(a - 5) chia hết cho 7 => (a - 5) + 21 chia hết cho 7 => a + 16 chia hết cho 7
(a - 6) chia hết cho 11 => (a - 6) + 22 chia hết cho 11 => a + 16 chia hết cho 11
=> a + 16 thuộc BC(5; 7; 11)
Mà BCNN(5; 7; 11) = 385
=> a + 16 thuộc B(385) = {0; 385; 770; ...}
=> a thuộc {-16; 369; 754;...}
Vì a là số tự nhiên nhỏ nhất
=> a = 369
b ) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.\)
Ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
.....................
\(\frac{1}{2012^2}=\frac{1}{2012.2012}< \frac{1}{2011.2012}\)
Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2012}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.< \frac{2011}{2012}\)
Mà \(\frac{2011}{2012}< 1\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\)
LM
Lê Minh Vũ
CTVHS
12 tháng 9 2021
\(b)\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)
\(< \)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)
\(< \)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(< \)\(1-\frac{1}{2012}\)\(=\frac{2011}{2012}< 1\)
Vậy Biểu thức \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< 1\)
PT
26 tháng 4 2016
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
26 tháng 4 2016
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
M
27 tháng 2 2016
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{2011^2}<\frac{1}{2010.2011};\frac{1}{2012^2}<\frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010+2011}+\frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2012}\)
Vì \(\frac{1}{2012}>0\) => \(\frac{1}{1}-\frac{1}{2012}<1\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<1\)
5 tháng 7 2018
Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};..........;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{2011.2012}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
5 tháng 7 2018
ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};...;\frac{1}{2011^2}< \frac{1}{2010.2011};\)\(\frac{1}{2012^2}< \frac{1}{2011.2012}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\left(đpcm\right)\)
PD
31 tháng 3 2019
Ta có
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)
= 1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2011}-\frac{1}{2012}\)
=1-\(\frac{1}{2012}\)=\(\frac{2011}{2012}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2012^2}< 1\)
Có:\(\frac{1}{2}<\frac{2}{3}\)
\(\frac{3}{4}<\frac{4}{5}\)
\(\frac{5}{6}<\frac{6}{7}\)
..............
\(\frac{2011}{2012}<\frac{2012}{2013}\)=>A<\(\frac{2}{3}.\frac{4}{5}.....\frac{2012}{2013}\)
=> A2<\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{2011}{2012}.\frac{2012}{2013}\)
=> A2<\(\frac{1.2.3.4.....2011.2012}{2.3.4.5.....2012.2013}\)
=>A2<\(\frac{1}{2013}\)=>DPCM