Với mọi N >= 2, hãy so sánh:
A = 1/12 + 1/32 + 1/42 + ... + 1/n2 với B = 1
K
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ND
2
MT
16 tháng 7 2015
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=\frac{1}{1}-\frac{1}{n}=1-\frac{1}{n}<1\left(\text{vì n}\ge2\text{ hay n dương}\right)\)
Vậy A<1
K
6 tháng 8 2018
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NX
0
8 tháng 12 2015
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
MA
1
21 tháng 8 2016
Ta có (21 -1)(21 + 1) = 22 - 1
(22 - 1)(22 + 1) = 24 - 1
tương tự như vậy ta sẽ có (2 -1)A = 232 - 1
vậy A < 232
AH
Akai Haruma
Giáo viên
7 tháng 12 2023
Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
\(A=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\\ =1+\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{n^2}\right)>1=B\)
Vậy \(A>B\)