The given sequence is:
$9, 24, 45, 72, 105, \dots$

To find the general formula for the nth term of this sequence, let's analyze the pattern.

1. First differences:
   $24 - 9 = 15$
   $45 - 24 = 21$
   $72 - 45 = 27$
   $105 - 72 = 33$

The first differences are increasing by 6 each time, indicating a quadratic sequence.

2. Second differences:
   $21 - 15 = 6$
   $27 - 21 = 6$
   $33 - 27 = 6$

Since the second differences are constant, this is a quadratic sequence. The general form of the nth term for a quadratic sequence is:

$T_n = an^2 + bn + c$

We now need to determine the coefficients $a$, $b$, and $c$.

3. Setting up equations:
   We know the first few terms of the sequence:

• $T_1 = 9$
• $T_2 = 24$
• $T_3 = 45$

Substitute these into the general form:

$T_1 = a(1)^2 + b(1) + c = 9 \quad \Rightarrow \quad a + b + c = 9$ $T_2 = a(2)^2 + b(2) + c = 24 \quad \Rightarrow \quad 4a + 2b + c = 24$ $T_3 = a(3)^2 + b(3) + c = 45 \quad \Rightarrow \quad 9a + 3b + c = 45$

We now have a system of three equations:

1. $a + b + c = 9$
2. $4a + 2b + c = 24$
3. ( 9a