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

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

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

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

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

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

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

$\Rightarrow A$ ko phải số tự nhiên

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1-122+1-132+1-142+....+1-120232"" id="MathJax-Element-2-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=1−122+1−132+1−1(2+....+1)120232�=1-122+1-132+1-142+....+1-1202321+12+13+...+122023−1

2022-(122+132+142+...+120232)"" id="MathJax-Element-3-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=2022−(122+132+142+...+120232)�=2022-(122+132+142+...+120232)A

k

làm 1 bài thôi có được không.

#ha le ha ban trả lời câu 2,3,4 giúp minh với

A=481/280

Bài 1:

A = 5 + 5^2 + 5^3 +...+ 5^8

A = ( 5 + 5^2 ) + ( 5^3 + 5^4 ) +...+ (5^7 +5^8)

A = 1.(5+5^2) + 5^2 . (5+5^2) +...+ 5^6.(5+5^2)

A = 1.30 + 5^2.30 +...+ 5^6.30

A = (1+5^2+...+5^6).30

Vì trong 2 thừa số có 1 thừa số chia hết cho 30 nên A chia hết cho 30

B = 3 + 3^3 + 3^5 +...+ 3^29

B = (3+ 3^3 +3^5)+...+(3^25+3^27+3^29)

B = 1.(3+3^3+3^5)+...+3^24. (3+3^3+3^5)

B = 1.273+...+3^24.273

B = (1+...+3^24).273

Vì trong 2 thừa số có 1 thừa số chia hết cho 273 nên B chia hết cho 273

A=5+5^2+5^3+...+5^20

=(5+5^2)+(5^3+5^4)+...+(5^19+5^20)

=(5+5^2)+5^2(5+5^2)+...5^18(5+5^2)

=30+5^2.30+5^4.30+5^6.30+..+5^18.30

=30(1+5^2+5^4+5^6+..+5^18)(chia hết cho 30)

Vậy A là bội của 30