Please solve in English

We need to show that the formula holds for $n=1$.
$M^1 = \begin{pmatrix} 3^1 & 3^1-2^1 \\ 0 & 2^1 \end{pmatrix} = \begin{pmatrix} 3 & 3-2 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$.
This matches the given matrix $M$. So, the base case holds.

Assume that the formula holds for some positive integer $k$. That is, assume:

We need to prove that the formula holds for $n=k+1$, i.e., $M^{k+1} = \begin{pmatrix} 3^{k+1} & 3^{k+1}-2^{k+1} \\ 0 & 2^{k+1} \end{pmatrix}$.
We know that $M^{k+1} = M^k \cdot M$.
Substitute the inductive hypothesis for $M^k$ and the given matrix $M$:

Perform the matrix multiplication:
The element in the first row, first column is:
$(3^k)(3) + (3^k-2^k)(0) = 3^{k+1}$
The element in the first row, second column is:
$(3^k)(1) + (3^k-2^k)(2) = 3^k + 2 \cdot 3^k - 2 \cdot 2^k = 3^k + 2 \cdot 3^k - 2^{k+1} = 3 \cdot 3^k - 2^{k+1} = 3^{k+1} - 2^{k+1}$
The element in the second row, first column is:
$(0)(3) + (2^k)(0) = 0$
The element in the second row, second column is:
$(0)(1) + (2^k)(2) = 2^{k+1}$
So, we get:

This matches the formula for $n=k+1$.
By the principle of mathematical induction, the formula is true for all positive integers $n$.

Given $y = \frac{x^3}{6} + \frac{1}{2x} = \frac{1}{6}x^3 + \frac{1}{2}x^{-1}$.
Differentiate with respect to $x$:

Now, add 1:

Notice that this expression is a perfect square:

The arc length $L$ of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the formula:

Substitute the expression from Step 2:

Evaluate at the limits:

The length of the curve is $\boxed{\frac{17}{12}}$.