It is Problem 1 from the column "Topics of the written maturity examination in high schools with a mathematical and physical profile", in the magazine MATHEMATYKA, No. 6/1986, page 363.
In translate, thanks to Mr. DeepL:
"Prove that if n \in \mathbb{N}, then 6^{2n}+3^{n+2}+3^n is divisible by 11."
We will note a \mid b if the number a divides the number b, respectively b\; \vdots \; a the equivalent statement that the number b is divisible by the number a.
SOLUTION 1 CiP
Solution by Mathematical Induction. Let c_n \overset{def}{=}6^{2n}+3^{n+2}+3^n.
For n=0,\;c_0=6^0+3^2+3^0=11, so c_0 \;\vdots \;11.
We assume the statement true for n=k,\;c_k \;\vdots \;11. Then
c_{k+1}=6^{2(k+1)}=3^{(k+1)+2}+3^{k+1}=6^{2k+2}+3^{k+3}+3^{k+1}=6^{2k}\cdot 36+3^{k+3}+3^{k+1}=
=3(12 \cdot 6^{2k}+3^{k+2}+3^k)\;\underset{3^{k+2}+3^k=c_k-6^{2k}}{=}\;3(12\cdot 6^{2k}+c_k-6^{2k})=
=3(11\cdot 6^{2k}+c_k)\;\;\vdots\;\;11,
because in parenthesis all terms are divisible by 11. So the statement is also true for n=k+1.
By Mathematical_Induction, the statement is true for all n\in \mathbb{N}.
\blacksquare
SOLUTION 2 CiP
We write the given expression successively
6^{2n}+3^{n+2}+3^n=
=2^{2n}\cdot 3^{2n}+3^{n+2}+3^n=3^n\cdot(2^{2n}\cdot 3^n+3^2+1)=3^n(4^n\cdot 3^n+10)=
=3^n({12}^n+10)=3^n[(11+1)^n+10]=3^n[(11\cdot d +1)+10]=3^n(d+1)\cdot 11\;\vdots\;11.
We used that, in developing (11+1)^n with Newton's binomial formula, the terms, except the last one which are 1^n, are divisible by 11.
\blacksquare \;\blacksquare
REMARKS
1^R. We see in the image that Problem 2 is of the same kind.
Directly, using the formula (a+b)^n=a\cdot d+b^n, we have
c_n\;\overset{def}{=}2^{6n+1}+3^{2n+2}=2\cdot {64}^n+9 \cdot 9^n=2\cdot(66-2)^n+9\cdot (11-2)^n=
=2(66\cdot d_1+(-2)^n)+9(11\cdot d_2+(-2)^n)=11\cdot (12d_1+9d_2)+2\cdot (-2)^n+9 \cdot (-2)^n=
=11d_3+11 \cdot (-2)^n\;\;\vdots \;11.
In another way, through Mathematical Induction as in Solution 1, we can use the formulas
c_{n+1}=2^{6n+7}+3^{2n+4}=64 \cdot 2^{6n+1}+9\cdot3^{2n+2}=64c_n-55\cdot 3^{2n+2},
or,
c_{n+1}=9c_n+55\cdot 2^{6n+1}
whence we see that c_n\;\vdots 11\;\;\Rightarrow \;c_{n+1}\;\vdots 11.
end Rem. 1
2^R. The numbers in Problem 1 are in fact 6^{2n}+10\cdot 3^n so we have a whole class of problems of the type
\alpha \cdot a^{k\cdot n}+\beta \cdot b^{l\cdot n}\;\vdots \;p. \tag{1}
We just have to find suitable integers \alpha, \beta \in \mathbb{Z},\;k,l \in \mathbb{N} and (possibly prime) p so that (1) is true for any n\in \mathbb{N}.
end Rem. 2
