Chứng minh rằng \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< 1\)
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Những câu hỏi liên quan
21 tháng 3 2023
1/2^2<1/1*2
1/3^2<1/2*3
...
1/10^2<1/9*10
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
=>A<9/10
=>A<1
H9
HT.Phong (9A5)
CTVHS
10 tháng 11 2023
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)
\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A< 1-\dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{9}{10}\)
\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))
15 tháng 1
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)
...
\(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)
Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}>\dfrac{90.9}{303}=\dfrac{3}{10}\)(1)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)(2)
Từ (1),(2) suy ra \(\dfrac{3}{10}< \dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
LT
19 tháng 4 2023
Nhận xét:
1/2 mũ 2 + 1/3 mũ 2 + 1/4 mũ 2 + ... + 1/10 mũ 2 <1/1.2 + 1/2.3 + 1/3.4 + ... + 1/ 9.10
Ta có:
A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/9 - 1/10
A = 1 - 1/10
A = 9/10
=> A < 1
Mà S < A
=> S < 1
DK
18 tháng 9 2022
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
MN
14 tháng 7 2021
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot2}-\dfrac{1}{2\cdot2\cdot2}+\dfrac{1}{2\cdot2\cdot2}-\dfrac{1}{2\cdot2\cdot2\cdot2}+.....+\dfrac{1}{2^{10}}\)
\(A=1-\dfrac{1}{2^{10}}\)
\(A+\dfrac{1}{2^{10}}=1-\dfrac{1}{2^{10}}+\dfrac{1}{2^{10}}=1\left(dpcm\right)\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$
$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$
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`.
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
cộng vế với vê sta đc
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}< 1\)
Vậy ta có đpcm