Chứng minh:
(2^1+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)=28^2-1
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16 tháng 5 2017
A=\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì\(10^8-1>10^8-3\)
\(\Rightarrow\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)
Vậy \(A< B\)
DT
0
NA
1
S
14 tháng 5 2017
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)
\(2^2A=\frac{2^2}{4^2}+\frac{2^2}{6^2}+\frac{2^2}{8^2}+...+\frac{2^2}{100^2}\)
\(4A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow4A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=> \(4A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=>\(4A< 1-\frac{1}{50}\)
=> 4A < 1
=> A < \(\frac{1}{4}\)(đpcm)
DV
0
DV
1
31 tháng 7 2015
bn rất tốt nhưng mk rất tiếc phải ns câu này
: mấy bn ấy qa OLM cổ chơi hết rùi
DV
5
RS
2 tháng 5 2018
Ta có: \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}< \frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(B< \frac{1}{2}\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{98\cdot100}\right)\)
\(B< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(B< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B< \frac{1}{4}-\frac{1}{200}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
31 tháng 7 2015
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}<\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2n-2\right).2n}\)
\(=\) \(\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right).\frac{1}{2}\)
=\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right).\frac{1}{2}\)
=\(\left(\frac{1}{2}-\frac{1}{2n}\right).\frac{1}{2}=\frac{1}{4}-\frac{1}{2n.2}<\frac{1}{4}\)
=> \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}<\frac{1}{4}\) điều phải chứng minh
DD
31 tháng 7 2015
Ta có 3.5<4.4=4^2
5.7<6^2
...
(2n-1)(2n+1)<4n^2
Do vậy 1/4^2+1/6^2+....+1/4n^2<1/3.5+1/5.7+....+1/(2n-1)(2n+1)
=1/2[1/3-1/5+1/5-.....+1/(2n-1)-1/(2n+1)]
=1/2(1/3-1/2n+1)
=1/6-1/2(2n+1)<1/4. Vậy ta có đpcm
DD
31 tháng 7 2015
Ta có 4^2>3.5
6^2>5.7
...
(2n-1)(2n+1)<4n^2
Do vậy 1/4^2+1/6^2+....+1/4n^2<1/3.5+1/5.7+...+1/(2n-1)(2n+1)
=1/2(1/3-1/5+1/5-...+1/2n-1-1/2n+1)
=1/2(1/3-1/2n+1)
=1/6-1/2(2n+1)<1/4 (đpcm
11 tháng 3 2021
vì: \(\dfrac{1}{4^2}< \dfrac{1}{4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{4}\)
........
\(\dfrac{1}{2020^2}< \dfrac{1}{4}\)
=> \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
Biến đổi vế trái ta có:
\(\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2-1\right)\left(2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^8-1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^{16}-1\right)\left(2^6+1\right)\)
=> Sai đề
\(VT=\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^1-1\right)\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^{16}-1\right)\left(2^6+1\right)\left(2^8+1\right)\)
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