\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+\dfrac{15}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{2023^2}\)

\(A=(1+1+1+...+1)-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{2023^2})\)

Tổng số hạng của 2 ngoặc trên bằng nhau và =(2023-2):1+1=2022(số hạng)

\(A=2022-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2})\)

Ta thấy:

\(0<\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)

Ta có

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2022.2023}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{2022}-\dfrac{1}{2023}\)

\(=1-\dfrac{1}{2023}<1\)

Do đó,2021<A<2022 

Vậy giá trị của A không phải 1 số tự nhiên(đpcm)

Ko bt

A=223​+328​+4215​+...+2023220232−1​

\(A = \frac{2^{2} - 1}{2^{2}} + \frac{3^{2} - 1}{3^{2}} + \frac{4^{2} - 1}{4^{2}} + . . . + \frac{202 3^{2} - 1}{202 3^{2}}\)

\(A = 1 - \frac{1}{2^{2}} + 1 - \frac{1}{3^{2}} + 1 - \frac{1}{4^{2}} + . . . + 1 - \frac{1}{202 3^{2}}\)

\(A = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . + \frac{1}{202 3^{2}} \left.\right)\)

Tổng số hạng của 2 ngoặc trên bằng nhau và =(2023-2):1+1=2022(số hạng)

\(A = 2022 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{202 3^{2}} \left.\right)\)

Ta thấy:

\(0 < \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{202 3^{2}} < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . + \frac{1}{2022.2023}\)

Ta có

\(\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . + \frac{1}{2022.2023}\)

\(= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . + \frac{1}{2022} - \frac{1}{2023}\)

\(= 1 - \frac{1}{2023} < 1\)

Do đó,2021<A<2022 

Vậy giá trị của A không phải 1 số tự nhiên(đpcm)

bài làm

Ta có:

\(\frac{2024}{2023^{2} + k} = \frac{2023^{2} + 2023}{2023^{2} + k} = 1 + \frac{2023 - k}{2023^{2} + k}\)

Vậy

\(A = \sum_{k = 1}^{2023} \left(\right. 1 + \frac{2023 - k}{2023^{2} + k} \left.\right) = 2023 + \sum_{k = 1}^{2023} \frac{2023 - k}{2023^{2} + k}\)

Vì \(\frac{2023 - k}{2023^{2} + k} > 0\) khi \(k < 2023\), và bằng 0 khi \(k = 2023\), nên

\(2023 < A < 2024\)

Suy ra A ko phải là số tự nhiên

$A=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+....+\frac{{2023}^2-1}{{2023}^2}$

$A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+....+1-\frac{1}{{2023}^2}$

$A=2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$

$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}$

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2022.2023}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{2023}$

$\Rightarrow 0<\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2}<1-\frac{1}{2023}<1$

$\Rightarrow 2022-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2023}^2})$ko phải số tự nhiên

$\Rightarrow A$ ko phải số tự nhiên

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k

Sửa đề: \(A=1+2^2+2^4+...+2^{2022}\)

\(\Leftrightarrow4\cdot A=2^2+2^4+2^6+...+2^{2024}\)

=>\(4A-A=2^2+2^4+...+2^{2024}-1-2^2-...-2^{2022}\)

=>\(3A=2^{2024}-1\)

mà \(2\cdot B=2^{2024}\)

nên 3A và 2B là hai số tự nhiên liên tiếp

Sửa đề: \(A=\frac{3}{2^2}+\frac{8}{3^2}+\ldots+\frac{2023^2-1}{2023^2}\)

Ta có; \(A=\frac{3}{2^2}+\frac{8}{3^2}+\ldots+\frac{2023^2-1}{2023^2}\)

\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\cdots+\frac{2023^2-1}{2023^2}\)

\(=1-\frac{1}{2^2}+1-\frac{1}{3^2}+\cdots+1-\frac{1}{2023^2}=2022-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2023^2}\right)<2022\)

ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{2023^2}<\frac{1}{2022\cdot2023}=\frac{1}{2022}-\frac{1}{2023}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2023^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2022}-\frac{1}{2023}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2023^2}<1\)

=>\(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2023^2}\right)>-1\)

=>\(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2023^2}\right)+2022>-1+2022=2021\)

=>2021<A<2022

=>A không là số tự nhiên