PN
3
K
Khách
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BN
15 tháng 7 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}\right)+\left(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}\right)\)
Ta có:
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) (1)
Lại có:
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\) (2)
Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
PM
Cho A=\(\frac{1}{1x2}+\frac{1}{3x4}+\frac{1}{5x6}+....+\frac{1}{99x100}\)
Chứng minh rằng: 7/12<A<5/6
4
AM
1 tháng 7 2015
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)
và \(A
TM
1
TD
23 tháng 12 2020
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
Vậy\(A=\frac{19}{20}\)
LV
0
LV
4
LV
2
LV
3
LV
0
Cho:A=1/1x2+1/3x4+....+1/99x100
CMR:7/12<A<5/6
Ta có:
A= 1/1x2 +1/3x4 +1/5x6 +...+ 1/99x100
A= 1-1/2 + 1/3 - 1/4 + 1/5 -1/6 +...+ 1/99-1/100
A= 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +...+1/99 + 1/100 - 2.1/2 - 2.1/4 - ... - 2.1/98
A= 1 + ... + 1/100 - 1 - 1/2 - 1/3 - ... - 1/49
A= 1/51 + ... + 1/100
=> A < 1/51.25 = 25/51 < 25/30 = 5/6 => đpcm
Và : A > 25x1/75 + 25x1/100 = 7/12
Ta có:
A= 1/1x2 +1/3x4 +1/5x6 +...+ 1/99x100
A= 1-1/2 + 1/3 - 1/4 + 1/5 -1/6 +...+ 1/99-1/100
A= 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +...+1/99 + 1/100 - 2.1/2 - 2.1/4 - ... - 2.1/98
A= 1 + ... + 1/100 - 1 - 1/2 - 1/3 - ... - 1/49
A= 1/51 + ... + 1/100
=> A < 1/51.25 = 25/51 < 25/30 = 5/6 => đpcm
Và : A > 25x1/75 + 25x1/100 = 7/12