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 ?
 A: 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 $20\equiv1,18\equiv-1,50\equiv12\pmod{19}$
A: Because $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n=$$
$$=20(50^n-12^n)+38\cdot12^n$$ and since
$$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1}),$$
we are done!
A: Hint:
$
5^{2n+1}  2^{n+2} + 3^{n+2}  2^{2n+1} 
\\= 20\cdot 50^n + 18 \cdot 12^n 
\\= 19(50^n + 12^n)  + 50^n - 12^n 
\\= 19(50^n + 12^n)  + (2\cdot 19 +12)^n - 12^n
$
A: $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20(50)^n+18(12)^n\equiv-18(50)^n+18(12)^n\equiv-18(12)^n+18(12)^n\equiv0$$
A: $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}\equiv 5^{2n}2^n-3^n2^{2n}\pmod{19}\qquad (1)$$
since $5^12^2=20\equiv 1\pmod{19}$ and $3^22^1=18\equiv -1\pmod{19}$. Hence, $(1)$ yields the following 
$$2^n(5^{2n}-3^n2^n)=2^n(25^n-6^n)=50^n-12^n\equiv 12^n-12^n\pmod{19}\equiv 0\pmod{19}$$
since $50\equiv 12\pmod{19}$.
A: Without congruences, subtraction, or factoring a difference of common powers:
Inductive assumption:
For some $n$, there is a $k$ such that
$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1} = 19k $
i.e.
$20(50^n)+18(12^n) = 19k$
For this $n$, add $980(50^n) + 198(12^n)$ to both sides. 
The left-hand side will turn into the desired $(n+1)^{th}$ expression. The right-hand side is now
$19k+11[20(50^n)+18(12^n)]+38[20(50^n)]$
Here just use the inductive assumption again, and these terms will have a common factor. 
A: First let's rewrite the expression so as to better highlight innate arithmetical structure.
$\ \ \quad\qquad\begin{align} 
19\ \mid &\,\ 5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}\\[.2em]
\iff\, \bmod 19\!:\, &\ \ \ \ \color{#0a0}{20}\cdot 50^n\, \equiv\, - \color{#0a0}{18}\cdot 12^n
\end{align}$
Now the induction is extremely simple if we view it arithmetically $\!\bmod 19,\,$ where it amounts to  multiplying the base congruence $\,\color{#0a0}{20}\equiv \color{#0a0}{-18}\,$ by the $\color{#c00}n$'th power of $\,50\equiv 12.\,$ In induction format
$\ \ \quad\qquad\qquad\qquad\begin{align} 
\color{#0a0}{20}\, \ &\equiv\, \color{#0a0}{-18}\qquad\qquad\ {\rm i.e.}\ \ \ P(0)\\[.3em]
\color{#0a0}{20}\cdot 50^{\large\color{#c00} n} &\equiv\, 12^{\large\color{#c00} n}(\color{#0a0}{-18})\qquad\! {\rm i.e.}\ \ \ P(\color{#c00}n)\\
\times\,\qquad 50\ \ &\equiv\, 12\\[.2em]
\hline
\!\!\Longrightarrow\ \ 20\cdot 50^{\large\color{#c00}{n+1}} &\equiv\, 12^{\large\color{#c00}{n+1}}(-18)\ \ \ {\rm i.e.}\ \ \ P(\color{#c00}{n\!+\!1})^{\phantom{|^|}}\!\!\!\!\!
\end{align}$
The final congruence is the product  the two prior congruences using the Congruence Product Rule. In number theory we make such deductions by using  Congruence Product and Power Rules. The power rule encapsulates such arithmetical inductions for convenient reuse.  The inductive proofs in the other answers are in fact special cases of the proof of the Power Rule (e.g. see here where I highlight this in great detail).
We can simplify further using $\,\color{#0a0}{20\equiv 1,\ 18\equiv -1},\ 50\equiv 12\pmod{\!19}\,$ to obtain
$\qquad\qquad\qquad\qquad\qquad\begin{align}&\color{#0a0}{20}\cdot 50^n +  \color{#0a0}{18}\cdot 12^n\\[.2em]
\equiv\ &\ \ 1\cdot 12^n\  \color{#0a0}{-\ 1}\cdot  12^n\end{align}$
which effectively completely encapsulates the induction in the Congruence Power Rule.
Remark $ $ Even if you don't know congruences you can still take advantage of these arithmetical simplifications by using a version of the Product Rule in divisibility form as below
$$\qquad\qquad\begin{align} {\rm mod}\,\ m\!:\, A\equiv a,\, B\equiv b&\ \ \,\Longrightarrow\,\ \ AB\equiv ab\qquad\bf\text{Congruence Product Rule}\\[3pt]
m\mid A-a,\ B-b&\,\Rightarrow\, m\mid AB-ab\qquad\bf\,\text{Divisibility$\ $ Product Rule}\\[4pt] 
{\bf Proof}\quad (A-a)B+a(B&-b)\, = AB-ab\end{align}$$
