Lời giải:

$11.5^{2n}+2^{3n+2}+2^{3n+1}=11.25^n+8^n.4+8^n.2=11.25^n+6.8^n$

Vì $25\equiv 8\pmod {17}$

$\Rightarrow 11.5^{2n}+2^{3n+2}+2^{3n+1} =11.25^n+6.8^n\equiv 11.8^n+6.8^n\equiv 17.8^n\equiv 0\pmod {17}$

Hay $11.5^{2n}+2^{3n+2}+2^{3n+1}\vdots 17$

Hay $

\(11.5^{2n}+2^{3n+2}+2^{3n+1}\)

\(=17.5^{2n}-6.5^{2n}+2^{3n}.6\)

\(=17.5^{2n}-6\left(5^{2n}-2^{3n}\right)\)

\(=17.5^{2n}-6\left(25^n-8^n\right)\)

Có \(17.5^{2n}⋮17\)

\(25^n-18^n⋮\left(25-18\right)⋮17\left(với\forall n\right)\)

\(\RightarrowĐpcm\)

11.5²ⁿ + 2³ⁿ⁺² + 2³ⁿ⁺¹ 

= 11.25ⁿ + 4.2³ⁿ + 2.2³ⁿ

= 17.25ⁿ - 6.25ⁿ + 2.2³ⁿ.(2+1)

= 17.25ⁿ -  6.25ⁿ + 6.2³ⁿ

= 17.25ⁿ - 6.(25ⁿ - 2³ⁿ)

= 17.25ⁿ - 6.(25ⁿ - 8ⁿ)

mà 25 - 8 = 17 chia hết cho 17

=> 25ⁿ - 8ⁿ chia hết cho 17

=> 17.25ⁿ - 6.(25ⁿ - 8ⁿ) chia hết cho 17

=> đpcm

Đặt \(A=11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

\(A=11\cdot25^n+8^n\cdot4+8^n\cdot2\)

\(A=17\cdot25^2-6\left(25^n-8^n\right)\)

\(A=17\cdot25^n-6\left(25-8\right)\left(25^{n-1}+25^{n-2}\cdot8+..........+8^{n-2}\cdot25+8^{n-1}\right)\)\(A=17\cdot25^n-17\cdot6\cdot\left(25^{n-1}+25^{n-2}\cdot8+..........+8^{n-2}\cdot25+8^{n-1}\right)\)\(\Rightarrow A⋮17\)

Sửa đề: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

Ta có: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

\(=11\cdot25^n+8^n\cdot4+8^n\cdot2\)

\(=11\cdot25^n+6\cdot8^n\)

Vì \(25\equiv8\)(mod 17)

nên \(11\cdot25^n+6\cdot8^n\equiv11\cdot8^n+6\cdot8^n\equiv17\cdot8^n\equiv0\)(mod 17)

hay \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}⋮17\)(đpcm)