Are they ugly boxes or matrices?

I'll prove by induction that:

$\large \displaystyle P^n = \begin{bmatrix} 1 & 0 & 0 \\ 4n & 1 & 0 \\ 8(n^2+n) & 4n & 1 \end{bmatrix}$

First the base case:

$\large \displaystyle P^1 = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} = P$

Which holds. Then the general case:

$\large \displaystyle P^n \cdot P = \begin{bmatrix} 1 & 0 & 0 \\ 4n & 1 & 0 \\ 8(n^2+n) & 4n & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 4n + 4 & 1 & 0 \\ 8(n^2+n) + 16n + 16 & 4n + 4 & 1 \end{bmatrix}$

$\large \displaystyle P^n \cdot P = \begin{bmatrix} 1 & 0 & 0 \\ 4(n + 1) & 1 & 0 \\ 8[(n+1)^2+(n+1)] & 4(n + 1) & 1 \end{bmatrix} = P^{n+1}$

This proves the induction.

Then:

$\large \displaystyle P^{50} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 20400 & 200 & 1 \end{bmatrix}$

$\color{#20A900} \boxed{ \large \displaystyle Q = P^{50} - I = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 20400 & 200 & 0 \end{bmatrix} }$

So:

$\color{#3D99F6} \large \displaystyle q_{31} = 20400, q_{32} = q_{21} = 200 \rightarrow \boxed{ \large \displaystyle \frac{q_{31}+q_{32}}{q_{21}} = 103}$ .

In fact this general ratio for $Q = P^n - I$ will be $2n+3$ .

P = \(\begin{bmatrix} 1 & 0 &0 \\ 4&1 & 0\\
16& 4 &1
\end{bmatrix} = \begin{bmatrix} 0 & 0 &0 \\ 4 &0 &0 \\ 16&4 &0 \end{bmatrix}

+

\begin{bmatrix} 1 &0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}\)

Let A = $\begin{bmatrix} 0 & 0 &0 \\ 4 &0 &0 \\ 16&4 &0 \end{bmatrix}$

$A^2$ = $\begin{bmatrix} 0 & 0 &0 \\ 0& 0 & 0\\ 16 & 0 &0 \end{bmatrix}$

$A^3$ = $\begin{bmatrix} 0 & 0 &0 \\ 0& 0 & 0\\ 0& 0 &0 \end{bmatrix}$

$A^n$ is also a null matrix for all $n \geq 3$

So

$P^{50}$ = $(I+A)^{50}$

Now Binomial expansion

$(I+A)^{50}$ = I + 50A + $\frac{50 * 49*A^2}{2}$ + 0

Here we get the value of $P^{50}$ - I

Now $\frac{q31+q32}{q21}$ = $\frac{16(50+25*49) +50*4}{50*4}$

= $\boxed{103}$