1. The problem is to find the $n^{th}$ term rule for the quadratic sequence: -4, -1, 4, 11, 20, ... 2. Calculate the first differences: $-1 - (-4) = 3$ $4 - (-1) = 5$ $11 - 4 = 7$ $20 - 11 = 9$ 3. Calculate the second differences: $5 - 3 = 2$ $7 - 5 = 2$ $9 - 7 = 2$ 4. Since the second differences are constant and equal to 2, the sequence is quadratic. 5. The general form of the $n^{th}$ term for a quadratic sequence is: $$a n^2 + b n + c$$ 6. The second difference equals $2a$, so: $$2a = 2 \Rightarrow a = 1$$ 7. Using the first term $T_1 = -4$: $$a(1)^2 + b(1) + c = -4 \Rightarrow 1 + b + c = -4 \Rightarrow b + c = -5$$ 8. Using the second term $T_2 = -1$: $$a(2)^2 + b(2) + c = -1 \Rightarrow 4 + 2b + c = -1 \Rightarrow 2b + c = -5$$ 9. Subtract the equation from step 7 from step 8: $$(2b + c) - (b + c) = -5 - (-5) \Rightarrow b = 0$$ 10. Substitute $b=0$ into the equation $b + c = -5$: $$0 + c = -5 \Rightarrow c = -5$$ 11. Therefore, the $n^{th}$ term rule is: $$T_n = n^2 - 5$$