To solve the problem step by step, we will follow the reasoning laid out in the video transcript and derive the solution systematically.
### Step 1: Understanding the Problem
We need to select 3 numbers from the set {1, 2, 3, ..., 2n + 1} such that they are in arithmetic progression (AP). The total number of ways to do this is given as 441.
### Step 2: Identifying the Conditions for AP
For three numbers \( a, b, c \) to be in AP, they must satisfy the condition \( a + c = 2b \). This implies that the sum of the first and last number must be even, which means both numbers must either be odd or even.
### Step 3: Counting Odd and Even Numbers
- The odd numbers in the set {1, 2, ..., 2n + 1} are: 1, 3, 5, ..., 2n + 1. There are \( n + 1 \) odd numbers.
- The even numbers in the set are: 2, 4, 6, ..., 2n. There are \( n \) even numbers.
### Step 4: Counting the Ways to Choose Numbers
The number of ways to choose 3 numbers in AP can be calculated by choosing:
1. 2 odd numbers from \( n + 1 \) odd numbers.
2. 2 even numbers from \( n \) even numbers.
Thus, the total number of ways to select the numbers is given by:
\[
\binom{n+1}{2} + \binom{n}{2} = 441
\]
### Step 5: Expanding the Combinations
Using the formula for combinations, we can expand the above expression:
\[
\binom{n+1}{2} = \frac{(n+1)n}{2}
\]
\[
\binom{n}{2} = \frac{n(n-1)}{2}
\]
Substituting these into our equation gives:
\[
\frac{(n+1)n}{2} + \frac{n(n-1)}{2} = 441
\]
### Step 6: Simplifying the Equation
Combining the terms:
\[
\frac{(n+1)n + n(n-1)}{2} = 441
\]
\[
\frac{n^2 + n + n^2 - n}{2} = 441
\]
\[
\frac{2n^2}{2} = 441
\]
\[
n^2 = 441
\]
### Step 7: Solving for \( n \)
Taking the square root:
\[
n = 21
\]
### Step 8: Finding the Sum of Divisors of \( n \)
Now we need to find the sum of the divisors of \( n = 21 \).
- The prime factorization of 21 is \( 3^1 \times 7^1 \).
- The formula for the sum of divisors \( \sigma(n) \) is given by:
\[
\sigma(n) = (1 + p_1 + p_1^2 + \ldots)(1 + p_2 + p_2^2 + \ldots)
\]
For \( n = 21 \):
\[
\sigma(21) = (1 + 3)(1 + 7) = 4 \times 8 = 32
\]
### Final Answer
The sum of the divisors of \( n \) is \( \boxed{32} \).
