Cho tổng : S=1/2^2 + 1/3^2 + ... + 1/99^2
Chứng minh 49/100< S <1
trl nhanh giúp mik vs ah
K
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NC
2
7 tháng 5 2017
ta có:1/2.2+1/3.3+....+1/99.99>1/2.3+1/3.4+1/4.5+...1/99.100=1/2-1/3+1/3-1/4+...+1/99-1/100=1/2-1/100=49/100
=> S>49/100
^_^
7 tháng 5 2017
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(\Rightarrow\frac{49}{100}< S\)
RS
0
TN
1
10 tháng 4 2017
VÌ \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...........;\frac{1}{99^2}=\frac{1}{99\cdot99}< \frac{1}{99\cdot100}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)\(\Rightarrow S< 1\)
VÌ \(\frac{1}{2\cdot3}< \frac{1}{2\cdot2};.....;\frac{1}{98\cdot99}< \frac{1}{99\cdot99}\)
\(\Rightarrow\)\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{98\cdot99}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}< S\)
\(\Rightarrow\frac{49}{100}< S< 1\)
\(K\)\(mk\)\(nha\)
18 tháng 7 2019
Ta có :
S= 1/51 +1/52 +..+1/100
Vì 1/51>1/52>...>1/100
=> S >1/100 * 50 =1/2 (1)
Vì 1/100 <1/99<...<1/51<1/50
=> S < 1/50 * 50=1 (2)
Từ (1),(2) => 1/2 < S<1
P=1/2^2+1/2^3+...+1/2^2018
2P=1/2 +1/2^2 +...+1/2^2017
=> 2P-P= (1/2 +1/2^2 +...+1/2^2017)-(1/2^2+1/2^3+...+1/2^2018 )
=> P=1/2 -1/2^2018 <1/2 <3/4
K
18 tháng 7 2019
Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{100}.50=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Ta có \(\frac{1}{51}< \frac{1}{50};\frac{1}{52}< \frac{1}{50};...;\frac{1}{100}< \frac{1}{50}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{50}.50=1\)
\(\Rightarrow S< 1\)
10 tháng 6 2018
Ta có :
1002 > 99 . 100
1012 > 100 . 101
..............
2002 > 199. 200
=> A < \(\frac{1}{99.100}+\frac{1}{100.101}+...+\frac{1}{199.200}=\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...+\frac{1}{199}-\frac{1}{200}\)
=> A < \(\frac{1}{99}-\frac{1}{200}< \frac{1}{99}\) \(\left(1\right)\)
Tương tự ta có :
A > \(\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{200.201}\)
=> A > \(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{200}-\frac{1}{201}\)
=> A > \(\frac{1}{100}-\frac{1}{201}>\frac{1}{100}-\frac{1}{200}\)
=> A > \(\frac{1}{200}\) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)Ta có :
\(\frac{1}{200}< A< \frac{1}{99}\)
=> ĐPCM
YN
0
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{99^2}< \dfrac{1}{98\cdot99}=\dfrac{1}{98}-\dfrac{1}{99}\)
Do đó: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
=>\(S< 1-\dfrac{1}{99}\)
=>S<1
\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{99^2}>\dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(S>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(S>\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
Do đó: \(\dfrac{49}{100}< S< 1\)