\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

Đặt A = 3¹ + 3² + 3³ + 3⁴ + ... + 3⁹⁹ + 3¹⁰⁰

= (3¹ + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)

= 3.(1 + 3) + 3³.(1 + 3) + ... + 3⁹⁹.(1 + 3)

= 3.4 + 3³.4 + ... + 3⁹⁹.4

= 4.(3 + 3³ + ... + 3⁹⁹) ⋮ 4

Vậy A ⋮ 4

.

\(B=3^1+3^2+3^3+...+3^{300}\\=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{299}+3^{300})\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+...+3^{299}\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{299}\cdot4\\=4\cdot(3+3^3+3^5+...+3^{299})\)

Vì \(4\cdot(3+3^3+3^5+...+3^{299})\vdots2\)

nên \(B\vdots2\)

B=(3+3²)+(3³+3⁴)+...+(3²⁹⁹+3³⁰⁰)

B=3(1+3)+3³(1+3)+...+3²⁹⁹(1+3)

B=3.4+3³.4+...+3^299.4

B=4(3+3³+...+3²⁹⁹) chia hết cho 2 vì 4 chia hết cho 2

vậy B chia hết cho 2

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

Ta có: \(\frac{1}{31}<\frac{1}{30};\frac{1}{32}<\frac{1}{30};...;\frac{1}{40}<\frac{1}{30}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}<\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (1)

Ta có: \(\frac{1}{41}<\frac{1}{40};\frac{1}{42}<\frac{1}{40};\ldots;\frac{1}{50}<\frac{1}{40}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}<\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (2)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{60}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{60}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{10}{50}=\frac15\) (3)

Từ (1),(2),(3) suy ra \(\left(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{60}\right)<\frac13+\frac14+\frac15\)

=>\(A<\frac{47}{60}<\frac{48}{60}=\frac45\)

\(B=3^0+3^1+3^2...+3^{100}\)

\(=3^0\times\left(1+3^1+3^2\right)+3^3\times\left(1+3^1+3^2\right)+...+3^{98}\times\left(1+3^1+3^2\right)\)

\(=3^0\times13+3^3\times13+...+3^{98}\times13\)

\(=13\times\left(3^0+3^3+...+3^{98}\right)⋮13\)

$B=3^0+3^1+3^2...+3^{100}$

$=3^0\times \left(1+3^1+3^2\right)+3^3\times \left(1+3^1+3^2\right)+...+3^{98}\times \left(1+3^1+3^2\right)$

$=3^0\times 13+3^3\times 13+...+3^{98}\times 13$

$=13$

\(A=3+3^2+3^3+...+3^{60}\)

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{57}+3^{58}+3^{59}+3^{60}\right)\)

\(\Rightarrow A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{57}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(3+3^5+...+3^{57}\right)\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=40\left(3+3^5+...+3^{57}\right)⋮40\)

$A=3+3^2+3^3+...+3^{60}$

$\Rightarrow A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{57}+3^{58}+3^{59}+3^{60}\right)$

$\Rightarrow A$