1. **State the problem:** We need to find the formula for the nth term of the quadratic sequence: 2, 3, 5, 8, 12. 2. **Identify the sequence type:** Since the sequence is quadratic, the nth term has the form $$a_n = An^2 + Bn + C$$ where $A$, $B$, and $C$ are constants. 3. **Find the first differences:** Calculate the differences between consecutive terms: $$3 - 2 = 1, \quad 5 - 3 = 2, \quad 8 - 5 = 3, \quad 12 - 8 = 4$$ 4. **Find the second differences:** Calculate the differences of the first differences: $$2 - 1 = 1, \quad 3 - 2 = 1, \quad 4 - 3 = 1$$ Since the second differences are constant and equal to 1, this confirms the sequence is quadratic. 5. **Use the formula for the second difference:** For a quadratic sequence, the second difference equals $2A$. So, $$2A = 1 \implies A = \frac{1}{2}$$ 6. **Set up equations using the first three terms:** For $n=1$, $$A(1)^2 + B(1) + C = 2 \implies \frac{1}{2} + B + C = 2$$ For $n=2$, $$A(2)^2 + B(2) + C = 3 \implies 4 \times \frac{1}{2} + 2B + C = 3 \implies 2 + 2B + C = 3$$ For $n=3$, $$A(3)^2 + B(3) + C = 5 \implies 9 \times \frac{1}{2} + 3B + C = 5 \implies \frac{9}{2} + 3B + C = 5$$ 7. **Simplify the equations:** Equation 1: $$\frac{1}{2} + B + C = 2 \implies B + C = 2 - \frac{1}{2} = \frac{3}{2}$$ Equation 2: $$2 + 2B + C = 3 \implies 2B + C = 1$$ Equation 3: $$\frac{9}{2} + 3B + C = 5 \implies 3B + C = 5 - \frac{9}{2} = \frac{1}{2}$$ 8. **Subtract equation 1 from equation 2:** $$(2B + C) - (B + C) = 1 - \frac{3}{2} \implies B = -\frac{1}{2}$$ 9. **Find $C$ using $B + C = \frac{3}{2}$:** $$-\frac{1}{2} + C = \frac{3}{2} \implies C = 2$$ 10. **Verify with equation 3:** $$3 \times -\frac{1}{2} + 2 = -\frac{3}{2} + 2 = \frac{1}{2}$$ which matches the right side. 11. **Write the nth term formula:** $$a_n = \frac{1}{2}n^2 - \frac{1}{2}n + 2$$ 12. **Optional simplification:** $$a_n = \frac{n^2 - n}{2} + 2$$ **Final answer:** $$\boxed{a_n = \frac{n^2 - n}{2} + 2}$$