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\(\dfrac{1}{2^2}\) +\(\dfrac{1}{3^2}^{^{ }}\)+...+\(\dfrac{1}{2023^2}\) < 1
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\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2023^2}\) chứng tỏ B<1
1
22 tháng 3 2023
Có : `1/2^2<1/(1*2)`
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\ ...\\ \dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023} \)
nên \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\\ =1-\dfrac{1}{2023}=\dfrac{2022}{2023}< 1\\ \Rightarrow B< 1\left(đpcm\right)\)
5 tháng 6 2023
\(A=\dfrac{2}{3}+\dfrac{2}{3^2}+\dfrac{2}{3^3}+....+\dfrac{2}{3^{2023}}\)
\(3A=2+\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2022}}\)
\(3A-A=\left(2+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{2}{3^{2022}}\right)-\left(\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2023}}\right)\)
\(2A=2-\dfrac{2}{3^{2023}}\)
\(A=\left(2-\dfrac{2}{3^{2023}}\right)\times\dfrac{1}{2}\)
\(A=2\times\dfrac{1}{2}-\dfrac{2}{3^{2023}}\times\dfrac{1}{2}\)
\(A=1-\dfrac{1}{3^{2023}}\)
=> \(A< 1\left(đpcm\right)\)
VD
16 tháng 4 2023
=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022
=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023
=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023
=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022
=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021) - 1/4^2022 - 2023/4^2022 + 2023/4^2023
=> 9S = 4 - 1/4^2022 - 2023/4^2022 + 2023/4^2023
= 4- 2024/4^2022 + 2023/4^2023
Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0
=> 9S < 4 < 9/2
=> S < 1/2 (đpcm)
LT
Lương Thị Vân Anh
CTVHS
16 tháng 4 2023
Ta có S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\)
4S = \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\)
4S - S = ( \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\) ) - ( \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\))
3S = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}-\dfrac{2023}{4^{2023}}\)
Đặt A = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\)
4A = 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)
4A - A = ( 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)) - ( 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\))
3A = 4 - \(\dfrac{1}{4^{2022}}\)
A = ( 4 - \(\dfrac{1}{4^{2022}}\)) : 3 = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\)
⇒ 3S = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)
S = ( \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)) : 3 = \(\dfrac{4}{9}-\dfrac{1}{4^{2022}\cdot3^2}-\dfrac{1}{4^{2023}\cdot3}< \dfrac{4}{9}< \dfrac{1}{2}\)
Vậy S < \(\dfrac{1}{2}\)
NN
0
OY
27 tháng 7 2021
a) Gọi ƯCLN(12n+1,30n+2) là d
12n+1⋮d ⇒ 60n+5⋮d
30n+2⋮d ⇒ 60n+4⋮d
(60n+5)-(60n+4)⋮d
1⋮d
Vậy \(\dfrac{12n+1}{30n+2}\) là ps tối giản
OY
27 tháng 7 2021
b) Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
NT
3
18 tháng 5 2022
\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{10^2}\)
Vì \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
\(A< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
Do đó \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{10}\Rightarrow A< \dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{10^2}< \dfrac{1}{2}\)
18 tháng 5 2022
`A = 1/3^2 + 1/4^2 + ... + 1/10^2`
Ta có:
`1/3^2 < 1/(2.3)`
`1/(4^2) < 1/(3.4)`
`...`
`1/(10^2) < 1/(9.10)`
`=> A < 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10 = 1/2 - 1/10 < 1/2`.
NN
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
\(2^2< 2.3\Rightarrow\dfrac{1}{2^2}>\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
Tương tự: \(\dfrac{1}{3^2}>\dfrac{1}{3}-\dfrac{1}{4}\) ; \(\dfrac{1}{4^2}>\dfrac{1}{4}-\dfrac{1}{5}\) ; ....; \(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)
Do đó:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{101}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2023}< 1\)