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Topic: Polygonal Numbers

• Objective: Finding the formula for the nth term of triangular numbers.

Triangular Numbers

• Triangular numbers are numbers that can form an equilateral triangle when represented as dots.

Sequence Example:

1 dot, 3 dots, 6 dots, 10 dots, 15 dots. These correspond to the sequence: $1, 3, 6, 10, 15$

Common difference:

The difference between consecutive terms is increasing:

• $3 - 1 = 2$
• $6 - 3 = 3$
• $10 - 6 = 4$
• $15 - 10 = 5$

Second difference:

• The second difference (difference of differences) is constant: $3 - 2 = 1$ $4 - 3 = 1$ $5 - 4 = 1$

Since the second difference is constant, this sequence is quadratic. Therefore, the formula for the nth term will be of the form: $T_n = an^2 + bn + c$

Step-by-step derivation of the nth-term formula:

1. Use the first few terms to create a system of equations:

   • $T_1 = 1: \ a(1)^2 + b(1) + c = 1 \Rightarrow a + b + c = 1$
   • $T_2 = 3: \ a(2)^2 + b(2) + c = 3 \Rightarrow 4a + 2b + c = 3$
   • $T_3 = 6: \ a(3)^2 + b(3) + c = 6 \Rightarrow 9a + 3b + c = 6$

2. Solve the system of equations:

   • From the first equation: $a + b + c = 1$ (Eq. 1)
   • From the second equation: $4a + 2b + c = 3$ (Eq. 2)
   • From the third equation: $9a + 3b + c = 6$ (Eq. 3)

   Subtract Eq. 1 from Eq. 2: $(4a + 2b + c) - (a + b + c) = 3 - 1$ $3a + b = 2 \quad \text{(Eq. 4)}$

   Subtract Eq. 2 from Eq. 3: $(9a + 3b + c) - (4a + 2b + c) = 6 - 3$ $5a + b = 3 \quad \text{(Eq. 5)}$

3. Solve for a and b: Subtract Eq. 4 from Eq. 5: $(5a + b) - (3a + b) = 3 - 2$ $2a = 1 \quad \Rightarrow \ a = \frac{1}{2}$

   Substitute $a = \frac{1}{2}$ into Eq. 4: [ 3\left(\frac{