<s>[INST] Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19. 
Prove that the expression 
  $$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
  is divisible by $19$.

I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II. 
Assume that 
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19. 
Now this is where I get lost, I try to "dismember" the expression to get 
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
 [/INST] Without using induction 
$$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n$$
$$\equiv1\cdot12^n+(-1)\cdot12^n\pmod{19}$$  as $20\equiv1,18\equiv-1,50\equiv12\pmod{19}$
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