Với mọi số tự nhiên n< hoặc bằng , so sánh A với 1 biết : A= 1\2² +1\3²+1\4² ..... +1\n²
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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}
K
6 tháng 8 2018
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4 tháng 1 2018
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )
27 tháng 12 2021
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)
NX
0
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Bài 2:
Với $n$ chẵn thì $n+4$ chẵn
$\Rightarrow (n+4)(n+7)$ là số chẵn
Với $n$ lẻ thì $n+7$ chẵn
$\Rightarrow (n+4)(n+7)$ là số chẵn
Vậy $(n+4)(n+7)$ chẵn với mọi số tự nhiên $n$ (đpcm)
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Bài 3:
a.
$101\vdots x-1$
$\Rightarrow x-1\in\left\{\pm 1; \pm 101\right\}$
$\Rightarrow x\in\left\{0; 2; 102; -100\right\}$
Vì $x\in\mathbb{N}$ nên $x=0, x=2$ hoặc $x=102$
b.
$a+3\vdots a+1$
$\Rightarrow (a+1)+2\vdots a+1$
$\Rightarrow 2\vdots a+1$
$\Rightarrow a+1\in\left\{\pm 1; \pm 2\right\}$
$\Rightarrow a\in\left\{0; -2; 1; -3\right\}$
IN
10 tháng 5 2016
1/ So sánh A với \(\frac{1}{4}\)
Có \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.........+\frac{1}{2014.2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-.......+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)
\(A=\frac{1}{1.2}-\frac{1}{2015.2016}=\frac{1}{2}-\frac{1}{2015.2016}\)
Vậy \(A>\frac{1}{4}\)