We are given a sequence of numbers: $1, \frac{2}{2^{2-1}}, \frac{3}{3^{2-2}}, \frac{4}{4^{2-3}}, ...$. We are to find a general formula for the $n$-th term of this sequence.

1. Problem Description

We are given a sequence of numbers: $1, \frac{2}{2^{2-1}}, \frac{3}{3^{2-2}}, \frac{4}{4^{2-3}}, ...$.

We are to find a general formula for the $n$-th term of this sequence.

2. Solution Steps

Let $a_n$ be the $n$-th term of the sequence.

We are given:

$a_1 = 1$

$a_2 = \frac{2}{2^{2-1}} = \frac{2}{2^1} = \frac{2}{2}$

$a_3 = \frac{3}{3^{2-2}} = \frac{3}{3^0} = \frac{3}{1}$

$a_4 = \frac{4}{4^{2-3}} = \frac{4}{4^{-1}} = \frac{4}{1/4} = 16$

We can rewrite $a_1$ as $a_1 = \frac{1}{1^{2-0}}$.

Based on the pattern, we can write the general term as:

$a_n = \frac{n}{n^{2-(n-1)}} = \frac{n}{n^{2-n+1}} = \frac{n}{n^{3-n}}$

$a_n = n^{1-(3-n)} = n^{1-3+n} = n^{n-2}$

Let's verify for the first few terms:

$a_1 = 1^{1-2} = 1^{-1} = 1$

$a_2 = 2^{2-2} = 2^0 = 1$

The original problem statement has $a_2 = \frac{2}{2^{2-1}} = \frac{2}{2} = 1$. My solution has $a_2=1$.

$a_3 = 3^{3-2} = 3^1 = 3$

The original problem statement has $a_3 = \frac{3}{3^{2-2}} = \frac{3}{3^0} = \frac{3}{1} = 3$. My solution has $a_3 = 3$.

$a_4 = 4^{4-2} = 4^2 = 16$

The original problem statement has $a_4 = \frac{4}{4^{2-3}} = \frac{4}{4^{-1}} = 4^2 = 16$. My solution has $a_4 = 16$.

Then we should find $a_n = n^{n-2}$ if $n>=2$ and $a_1=1$.

3. Final Answer

$a_n = n^{n-2}$ for $n > 1$ and $a_1 = 1$.

We can express this as:

$a_n =

\begin{cases}

1, & \text{if } n = 1 \\

n^{n-2}, & \text{if } n > 1

\end{cases}

$$

Or simply $a_n = n^{n-2}$ if we interpret $1^{1-2}=1^{-1}=1$.

So, $a_n = n^{n-2}$.

Final Answer: $n^{n-2}$