\(M=3^1+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)

\(M=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)...+3^{28}\left(1+3+3^2\right)\)

\(M=3.13+3^4.13...+3^{28}.13\)

\(M=13.\left(3+3^4...+3^{28}\right)⋮13\)

\(\Rightarrow dpcm\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{27}+2^{28}+2^{29}\right)\\ A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{27}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(1+2^3+...+2^{27}\right)\\ A=7\left(1+2^3+...+2^{27}\right)⋮7\)

\(S=2^1+2^2+2^3+2^4+2^5+2^6+..+2^{28}+2^{29}+2^{30}\) 

\(S=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{28}.\left(1+2+2^2\right)\) 

\(S=\left(1+2+2^2\right).\left(2+2^4+...+2^{28}\right)\) 

\(S=7.\left(2+2^4+...+2^{28}\right)\) 

⇒ \(S⋮7\)   ( điều phải chứng minh ) 

S=2¹+2²+2³+...+2³⁰

S=(2¹+2²+2³)+(2⁴+2⁵+2⁶)+...+(2²⁸+2²⁹+2³⁰)

S=7.2+7.2⁴+...+7.2²⁸

S=7.(2+2⁴+...+2²⁸)

⇒S⋮7

Ta có: \(S=2^1+2^2+2^3+...+2^{28}+2^{29}+2^{30}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{28}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+2^4+...+2^{28}\right)⋮7\)

\(a,A=5^1+5^2+...+5^{100}\)

\(\Rightarrow A=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)

\(\Rightarrow6\left(5+5^3+...+5^{99}\right)\)

\(\Rightarrow A⋮6\)

\(b,B=2+2^2+2^3+...+2^{28}+2^{29}+2^{30}\)

\(\Rightarrow B=2\left(1+2+2^2\right)+...+2^{28}\left(1+2+2^2\right)\)

\(\Rightarrow7\left(2+...+2^{28}\right)\)

\(\Rightarrow B⋮7\)

ban koko 

\(A=3+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{29}+3^{30}\right)\)

\(A=1\left(3+3^2\right)+3^2\left(3+3^2\right)+....+3^{28}\left(3+3^2\right)\)

\(A=\left(1+3^2+...+3^{28}\right)\left(3+3^2\right)\)

\(A=13\left(1+3^2+...+3^{28}\right)⋮13\left(đpcm\right)\)

a) (1+5+5²+5³+...5²⁹⁾chia hết cho 6

Đặt (1+5+5²+5³+...5²⁹) = A

\(A=\left(1+5\right)+\left(5^2+5^3\right)+\left(5^4+5^5\right)....+\left(5^{28}+5^{29}\right)\)

\(A=\left(1+5\right)+5^2\left(5+1\right)+5^4\left(5+1\right)+...+5^{28}\left(5+1\right)\)

\(A=6+5^2.6+5^4.6+...+5^{28}.6\)

Vậy A chia hết cho 6

b) (1+3+3^2+3^3+...+3^29) chia hết cho 13

Đặt B= (1+3+3^2+3^3+...+3^29)

\(B=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{27}+3^{28}+3^{29}\right)\)

\(B=13+3^3\left(1+3+3^2\right)+....+3^{27}\left(1+3+3^2\right)\)

\(B=13+3^3.13+....+3^{27}.13\)

Vậy B chia hết 13

Câu c,d tương tự.Chúc bạn học tốt

Ta có \(M=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)

\(=3\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{28}.\left(1+3+3^2\right)\)

\(=13\left(3+3^4+...+3^{28}\right)⋮13\Rightarrow M⋮13\)

M = 3¹ + 3² + 3³ +...+ 3²⁸ + 3²⁹ + 3³⁰

M = ( 3¹ + 3² + 3³) + ...+ ( 3²⁸ + 3²⁹ + 3³⁰ )

M = 3(1 + 3 + 3² ) +...+ 3²⁸( 1 + 3 + 3²)

M = 3 .13 +...+ 3²⁸.13

\(\Rightarrow M⋮13\)(đpcm)

   !!!

Ta có \(M=3^1+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)

\(=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)

\(=3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{28}.\left(1+3+3^2\right)\)

\(=3.13+3^4.13+...+3^{28}.13\)

\(=13.\left(3+3^4+..+3^{28}\right)\) chia hết cho 13.

Vậy M chia hết cho 13