A positive integer $K$ is given. Define the sequence $(a_n)$ by $a_1 = 1$ and $a_n$ is the $n$-th positive integer greater than $a_{n-1}$ which is congruent to $n$ modulo $K$.

[b](a)[/b] Find an explicit formula for $a_n$.

[b](b)[/b] What is the result if $K = 2$?
1. **Understanding the sequence definition:**
   - We start with \( a_1 = 1 \).
   - For \( n \geq 2 \), \( a_n \) is the \( n \)-th positive integer greater than \( a_{n-1} \) which is congruent to \( n \mod K \).

2. **Finding the explicit formula for \( a_n \):**
   - We need to find \( a_n \) such that \( a_n \equiv n \pmod{K} \) and \( a_n \) is the \( n \)-th such integer greater than \( a_{n-1} \).
   - Let's denote the \( n \)-th positive integer greater than \( a_{n-1} \) which is congruent to \( n \mod K \) as \( a_{n-1} + mK + n \), where \( m \) is an integer.

3. **Deriving the formula:**
   - Since \( a_n \) is the \( n \)-th such integer, we need to find \( m \) such that there are \( n \) integers of the form \( a_{n-1} + mK + n \) greater than \( a_{n-1} \).
   - The first such integer greater than \( a_{n-1} \) is \( a_{n-1} + K + n \).
   - The second such integer is \( a_{n-1} + 2K + n \).
   - Continuing this pattern, the \( n \)-th such integer is \( a_{n-1} + nK + n \).

4. **Simplifying the formula:**
   - Therefore, \( a_n = a_{n-1} + nK + n \).
   - We can rewrite this as \( a_n = a_{n-1} + n(K + 1) \).

5. **Finding the general term:**
   - We start with \( a_1 = 1 \).
   - For \( n = 2 \), \( a_2 = a_1 + 2(K + 1) = 1 + 2(K + 1) = 1 + 2K + 2 \).
   - For \( n = 3 \), \( a_3 = a_2 + 3(K + 1) = (1 + 2K + 2) + 3(K + 1) = 1 + 2K + 2 + 3K + 3 = 1 + 5K + 5 \).
   - Continuing this pattern, we get \( a_n = 1 + \sum_{i=1}^{n-1} i(K + 1) \).

6. **Summing the series:**
   - The sum of the first \( n-1 \) positive integers is \( \frac{(n-1)n}{2} \).
   - Therefore, \( a_n = 1 + (K + 1) \cdot \frac{(n-1)n}{2} \).

7. **Final formula:**
   - Simplifying, we get \( a_n = 1 + \frac{(K + 1)(n-1)n}{2} \).

8. **Special case when \( K = 2 \):**
   - Substituting \( K = 2 \) into the formula, we get \( a_n = 1 + \frac{(2 + 1)(n-1)n}{2} = 1 + \frac{3(n-1)n}{2} \).
   - Simplifying further, \( a_n = 1 + \frac{3n^2 - 3n}{2} = 1 + \frac{3n^2}{2} - \frac{3n}{2} \).
   - This simplifies to \( a_n = \frac{3n^2 - 3n + 2}{2} \).

The final answer is \( \boxed{ a_n = \frac{3n^2 - 3n + 2}{2} } \).