Chứng minh rằng:1/4^2+1/6^2+1/8^2+...+1/100^2<1/4
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3 tháng 9 2017
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
LT
1
1 tháng 11 2016
\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\)
\(2^2.A=1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\)
\(2^2.A-A=\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\right)\)
\(4.A-A=1-\frac{1}{2^{100}}< 1\)
\(3A< 1\)
\(\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)
NT
1
17 tháng 5 2017
Gọi dãy trên là A, Ta có:
1/52+1/62+1/72+...+1/1002 < 1/4.5+1/5.6+1/6.7+...+1/99.100
<=> 1/52+1/62+1/72+...+1/1002 < 1/4 - 1/100
<=> 1/52+1/62+1/72+...+1/1002 < 6/25
Mà 6/25 < 1/4 => A < 1/4
6/25 > 1/6 => A > 1/6
V ậ y: 1/6 < A < 1/4
21 tháng 8 2018
Ta thấy: k2 > (k - 1)(k + 1)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right).\dfrac{1}{2}\)
\(=\left(1-\dfrac{1}{101}\right).\dfrac{1}{2}\)
\(=\dfrac{100}{101}.\dfrac{1}{2}< 1.\dfrac{1}{2}=\dfrac{1}{2}\)
21 tháng 4 2016
ta có :\(\frac{1}{5^2}<\frac{1}{4.5}\)
\(\frac{1}{6^2}<\frac{1}{5.6}\)
\(\frac{1}{7^2}<\frac{1}{6.7}\)
.....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow A<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\) (1)
Ta có : \(\frac{1}{5.6}<\frac{1}{5^2}\)'
\(\frac{1}{6.7}<\frac{1}{6^2}\)
....\(\frac{1}{100.101}<\frac{1}{100^2}\)
\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\) <A
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{101}\) <A
\(\frac{1}{5}-\frac{1}{101}\) <A
mà \(\frac{96}{5.101}=\frac{96}{505}>\frac{96}{576}\)
hay \(A>\frac{1}{6}\) (2)
từ (1); và (2) suy ra \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+..+\frac{1}{100^2}<\frac{1}{4}\) (đpcm)
đây là cách dễ hiểu nhất nhé
NT
1
21 tháng 3 2023
1/4^2<1/3*4
1/5^2<1/4*5
...
1/100^2<1/99*100
=>A<1/3-1/4+1/4-1/5+...+1/99-1/100
=>A<1/3-1/100<1/3
DH
4
S
18 tháng 3 2017
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
\(=\dfrac{1}{2^2}\cdot\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1+1=2\)
\(\Rightarrow\dfrac{1}{2^2}.\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)
\(\Rightarrow dpcm\)
18 tháng 3 2017
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{99}{200}< \dfrac{1}{2}\)
ND
0
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)
\(2^2A=\frac{2^2}{4^2}+\frac{2^2}{6^2}+\frac{2^2}{8^2}+...+\frac{2^2}{100^2}\)
\(4A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
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)