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From the images you've uploaded, the first one illustrates **polygonal numbers** (square, pentagonal, hexagonal numbers), while the second image shows a whiteboard solving a similar type of number pattern problem—specifically related to triangular numbers using a **quadratic sequence**.

I'll now walk you through solving one of these sequences, particularly the pentagonal numbers (1, 5, 12, 22, 35) using a quadratic formula approach, as done in the whiteboard.

### Steps to find the nth term of the pentagonal sequence:

1. **Recognize that the sequence is quadratic**:  
   The nth term of a quadratic sequence is usually in the form:
   \[
   U_n = an^2 + bn + c
   \]
   where \( a \), \( b \), and \( c \) are constants to be determined, and \( n \) is the position in the sequence.

2. **Use the first few terms of the pentagonal sequence**:
   The pentagonal sequence given is: 1, 5, 12, 22, 35 for \( n = 1, 2, 3, 4, 5 \).

3. **Set up equations for the first few terms**:
   \[
   U_1 = a(1)^2 + b(1) + c = a + b + c = 1
   \]
   \[
   U_2 = a(2)^2 + b(2) + c = 4a + 2b + c = 5
   \]
   \[
   U_3 = a(3)^2 + b(3) + c = 9a + 3b + c = 12
   \]

4. **Solve the system of equations**:
   From the three equations, we get:
   1. \( a + b + c = 1 \)
   2. \( 4a + 2b + c = 5 \)
   3. \( 9a + 3b + c = 12 \)

   Subtracting equation (1) from (2):
   \[
   (4a + 2b + c) - (a + b + c) = 5 - 1
   \]
   \[
   3a + b = 4 \quad \text{(Equation 4)}
   \]

   Subtracting equation (2) from (3):
   \[
   (9a + 3b + c) - (4a + 2b + c) = 12 - 5
   \]
   \[
   5a + b = 7 \quad \text{(Equation 5)}
   \]

   Subtract equation (4) from (5):
   \[
   (5a + b) - (3a + b) = 7 - 4
   \]
   \[
   2a = 3 \quad \Rightarrow a = \frac{3}{2}
   \]

   Substituting \( a = \frac{3}{2} \) into equation (4):
   \[
   3\left(\frac{3}{2}\right) + b = 4
   \]
   \[
   \frac{9}{2} + b = 4 \quad \Rightarrow b = 4 - \frac{9}{2} = \frac{-1}{2}
   \]

   Substituting \( a = \frac{3}{2} \) and \( b = \frac{-1}{2} \) into equation (1):
   \[
   \frac{3}{2} + \frac{-1}{2} + c = 1
   \]
   \[
   c = 1 - 1 = 0
   \]

5. **Final formula**:
   The nth term formula for the pentagonal sequence is:
   \[
   U_n = \frac{3}{2}n^2 - \frac{1}{2}n
   \]
   This formula can now be used to generate the pentagonal numbers.

Would you like further details on any part of this solution or another example? Here are some related questions:

1. What is the general process for solving quadratic sequences?
2. How can this formula be adapted for other polygonal numbers?
3. Can this approach be used for cubic sequences?
4. How do triangular, pentagonal, and hexagonal numbers relate?
5. How do the differences between terms reveal the degree of a sequence?

**Tip**: Whenever solving quadratic sequences, always check the second difference between terms; if it's constant, you are dealing with a quadratic sequence.

Learn how to determine the nth term of polygonal number sequences, such as pentagonal and hexagonal numbers, using quadratic sequence formulas. This step-by-step guide covers essential algebraic concepts and theorems, suitable for students in Grades 8-10.

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