\(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+.............+3^{24}\left(3+2^3+3^5\right)\)

\(B=273+273\cdot3^6+.............+273\cdot3^{24}\)

\(B=273\left(1+3^6+.......+3^{24}\right)⋮273\)

\(A=\left(5+5^2\right)+\left(5+5^2\right)5^2+\left(5+5^2\right)5^4+\left(5+5^2\right)5^6+\left(5+5^2\right)5^8\)

\(A=30+30\cdot5^2+30\cdot5^4+30\cdot5^6+30\cdot5^8\)

\(A=30\left(1+5^2+5^4+5^6+5^8\right)⋮30\)

\(A=3+3^3+...+3^{29}=\left(3+3^3+3^5\right)+...+\left(3^{25}+3^{27}+3^{29}\right)=273+...+3^{25}.273=273.\left(1+...+3^{25}\right)\) chia hết cho 273

Vậy A là bội của 273

a) \(B=3+3^3+3^5+...+3^{29}\)

\(\Rightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\)

\(\Rightarrow B=\left(3+3^3+3^5\right)+...+3^{24}.\left(3+3^3+3^5\right)\)

\(\Rightarrow B=273+...+3^{24}.273\)

\(\Rightarrow B=273.\left(1+...+3^{24}\right)⋮273\)

Vậy B là bội của 273.

b) \(A=5+5^2+...+5^7+5^8\)

\(\Rightarrow A=\left(5+5^2\right)+...+\left(5^7+5^8\right)\)

\(\Rightarrow A=\left(5+5^2\right)+...+5^6.\left(5+5^2\right)\)

\(\Rightarrow A=30+...+5^6.30\)

\(\Rightarrow A=30.\left(1+...+5^6\right)⋮30\)

Vậy A là bội của 30.

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

\(A=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{25}+3^{27}+3^{29}\right)\)

     \(=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)\)

     \(=273+3^6.273+........+3^{24}.273\)

     \(=273\left(1+3^6+......+3^{24}\right)\)chia hết cho 273

Ta có:

273=3+3^3+3^5

A=(3+3^3+3^5)+(3^7+3^9+3^11)+...+(3^25+3^27+3^29)

A=1×(3+3^3+3^5)+3^6×(3+3^3+3^5)+...+3^24×(3+3^3+3^5)

A=1×273+3^6×273+...+3^24×273

A=(1+3^6+...+3^24)×273

Suy ra: A chia hết cho 273

a) \(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)=30+5^2.30+...+5^6.30\)

\(=30\left(1+5^2+...+5^6\right)⋮30\Rightarrowđpcm\)

b) \(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)=273+3^6.273+...+3^{24}.273\)

\(=273.\left(1+3^6+...+3^{24}\right)⋮273\Rightarrowđpcm\)

a: \(B=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)\)

\(=156\cdot5\cdot\left(1+5^4\right)\)

\(=780\left(1+5^4\right)⋮30\)

b: \(B=\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^2+3^5\right)\)

\(=273\cdot\left(1+...+3^{24}\right)⋮273\)

a)  \(\overline{aaaaaa}=a.111111=a.3.37037\) \(⋮\)\(37037\)

b)  Nhận thấy các hạng tử trong B  đều chia hết cho 3   =>  B chia hết cho 3

\(B=3+3^3+3^5+3^7+...+3^{2017}+3^{2019}+3^{2021}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+....+\left(3^{2017}+3^{2019}+3^{2021}\right)\)

\(=3\left(1+3^2+3^4\right)+3^7\left(1+3^2+3^4\right)+...+3^{2017}\left(1+3^2+3^4\right)\)

\(=\left(1+3^2+3^4\right)\left(3+3^7+...+3^{2017}\right)\)

\(=91\left(3+3^7+....+3^{2017}\right)\)\(⋮\)\(91\)

mà  (3;91) = 1

=>  B chia hết cho 273

B chia hết cho 273

Còn câu a thì mình không biết nhé, xin lỗi bạn.

Đề bài: Chứng tỏ rằng:

a) Giá trị của biểu thức A=5+5²+5³+...+5⁹ là bội của 31

Ta có: A=5+5²+5³+...+5⁹ 

            =(5 + 5² + 5³) + .... + (5⁶ + 5⁷ + 5⁹)

            = 5.31 + .... + 5⁶.31

            = 31.(5 + .... + 5⁶) là bội của 31