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\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\text{…}+\frac{1}{2^{n-1}}\)
\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+\text{…}+\frac{1}{2^{n-1}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\text{…}-\frac{1}{2^n}\)
\(A=1-\frac{1}{2^n}\)
Vậy A < 1 với n thuộc N*
a)Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
=> A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
=> A < 1 - 1/100
=> A < 99/100 < 1
b) \(A=\frac{n+3}{n-2}=\frac{n-2+5}{n-2}=1+\frac{5}{n-2}\)
Để A có giá trị nguyên <=> 5 chia hết cho n - 2
<=> n - 2 thuộc Ư(5) = {1; -1; 5; -5}
Lập bảng:
| n - 2 | 1 | -1 | 5 | -5 |
| n | 3 | 1 | 7 | -3 |
Vậy ....
Ta có: A = \(\frac{10^{2019}+1}{10^{2020}+1}\)
=> 10A = \(\frac{10^{2020}+10}{10^{2020}+1}=\frac{\left(10^{2020}+1\right)+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)
B = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10B = \(\frac{10^{2021}+10}{10^{2021}+1}=\frac{10^{2021}+1+9}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Do \(\frac{9}{10^{2020}+1}>\frac{9}{10^{2021}+1}\)=> \(1+\frac{9}{10^{2020}+1}>1+\frac{9}{10^{2021}+1}\)
=> 10A > 10B
=> A > B
Lời giải:
$a+a^2+a^3+...+a^{2n}=(a+a^2)+(a^3+a^4)+...+(a^{2n-1}+a^{2n})$
$=a(a+1)+a^3(a+1)+....+a^{2n-1}(a+1)$
$=(a+1)(a+a^3+....+a^{2n-1})\vdots a+1$
Ta có :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+............+\dfrac{1}{2^n}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..........+\dfrac{1}{2^{n-1}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{n-1}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^n}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^n}< 1\)
\(\Leftrightarrow A< 1\rightarrowđpcm\)
Vậy ...............
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{n-1}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{n-1}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}\right)\\ A=1-\dfrac{1}{2^n}< 1\)
Vậy \(A< 1\) với mọi n