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a) \(A=2+2^2+...+2^{2024}\)
\(2A=2^2+2^3+...+2^{2025}\)
\(2A-A=2^2+2^3+...+2^{2025}-2-2^2-...-2^{2024}\)
\(A=2^{2025}-2\)
b) \(2A+4=2n\)
\(\Rightarrow2\cdot\left(2^{2025}-2\right)+4=2n\)
\(\Rightarrow2^{2026}-4+4=2n\)
\(\Rightarrow2n=2^{2026}\)
\(\Rightarrow n=2^{2026}:2\)
\(\Rightarrow n=2^{2025}\)
c) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2023}+2^{2024}\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{2023}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{2023}\right)\)
d) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=2+\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2022}+2^{2023}+2^{2024}\right)\)
\(A=2+2^2\cdot7+2^5\cdot7+...+2^{2022}\cdot7\)
\(A=2+7\cdot\left(2^2+2^5+...+2^{2022}\right)\)
Mà: \(7\cdot\left(2^2+2^5+...+2^{2022}\right)\) ⋮ 7
⇒ A : 7 dư 2
Bài 1:
a) 02002 < 02023
b) 20220 = 20230
c) 549 < 5510
d) ( 4 + 5 )3 > 42 + 52
đ) 92 - 32 > ( 9 - 3 )2
Bài 2:
a) 32 x 43 - 32 + 333
= 9 x 64 - 9 + 333
= 576 - 9 + 333
= 567 + 333
= 900
b) 5 x 43 + 24 x 5 + 410
= 5 x 64 + 24 x 5 + 1
= 5 x ( 64 + 24 ) + 1
= 5 x 88 + 1
= 440 + 1
= 441
c) 23 x 42 + 32 x 5 - 40 x 12023
= 8 x 16 + 9 x 5 - 40 x 1
= 128 + 45 - 40
= 133
Bài 1 :
a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)
b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)
c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)
d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)
đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)
\(1+3+3^2+3^3+...+3^{2023}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2022}+3^{2023}\right)\)
\(=4+3^2\cdot\left(1+3\right)+...+3^{2022}\cdot\left(1+3\right)\)
\(=4+4\cdot3^2+4\cdot3^4+....+4\cdot3^{2022}\)
\(=4\cdot\left(1+3^2+3^4+...+3^{2022}\right)\)
Mà: \(4\cdot\left(1+3^2+3^4+...+3^{2022}\right)\) ⋮ 4
\(\Rightarrow1+3+3^2+3^3+....+3^{2023}\) ⋮ 4
2S = 1 + 3 + 3² + 3³ + ... + 3¹¹
⇒ 6S = 3 + 3² + 3³ + 3⁴ + ... + 3¹²
⇒ 4S = 6S - 2S = (3 + 3² + 3³ + 3⁴ + ... + 3¹²) - (1 + 3 + 3² + 3³ + ... + 3¹¹)
= 3¹² - 1
= 531440
⇒ S = 531440 : 4
= 132860 ⋮ 10
Vậy S ⋮ 10
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{19}\right)⋮7\)
Lời giải:
$A=9+2.3^2+2.3^3+2.3^4+...+2.3^{2023}$
$A-9=2(3^2+3^3+3^4+...+3^{2023})$
$3(A-9)=2(3^3+3^4+3^5+...+3^{2024})$
$\Rightarrow 3(A-9)-(A-9)=2(3^{2024}-3^2)$
$2(A-9)=2.3^{2024}-18$
$\Rightarrow 2A-18=2.3^{2024}-18$
$\Rightarrow A=3^{2024}\vdots 3^{2023}$ (đpcm)