Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mathjax tutorial?

(Alternative approach, without induction.) The following uses the notation $\,\mathcal{M}n\,$ to denote any integer multiple of $\,n\,$. The obvious rules apply $\,2 \cdot \mathcal{M}3 = \mathcal{M}6, \,\mathcal{M}3 \cdot \mathcal{M}3 = \mathcal{M}9\,$ etc.


*

*$\,4^k-1 = \mathcal{M}3\,$ for all $\,k \ge 0\,$, therefore:


$$\require{cancel}
\begin{align}
4^n - 1 &= (4-1)\cdot(4^{n-1}+\cdots+4+1) \\
 &=3 \cdot \big((4^{n-1}-1)+ \cdots + (4-1) + (\cancel{1} - \cancel{1}) + n\big) \tag{1}\\
 &= 3 \cdot (\mathcal{M}3 + n) \\
 &= \mathcal{M}9 + 3n
\end{align}
$$


*

*Using the previous result back in $\,(1)\,$:


$$
\begin{align}
\color{blue}{2 \cdot (4^n - 1)} &=6 \cdot \big((\mathcal{M}9+3(n-1))+ (\mathcal{M}9+3(n-2))+ \cdots + (\mathcal{M}9+3 \cdot 1) + n\big)\\
 &= \mathcal{M}54 + 18\big((n-1)+(n-2)+\cdots+1 \big) + 6n \\
 &= \mathcal{M}54 + 18 \cdot \frac{n(n-1)}{2}+6n \\
 &= \color{blue}{\mathcal{M}54 + 9 n^2 - 3n} \tag{2}
\end{align}
$$
It then follows straight from $\,(2)\,$ that: $$2^{2n+1} - 9n^2 + 3n -2 = \color{blue}{2 \cdot (4^n -1)} -9n^2+3n=\color{blue}{\mathcal{M}54 + \cancel{9 n^2} - \bcancel{3n}} - \cancel{9n^2} + \bcancel{3n}$$
