1. **State the problem:** We are given the first five terms of a quadratic sequence: 1, 6, 13, 22, 33. We need to find an expression for the nth term of the sequence. 2. **Recall the formula for the nth term of a quadratic sequence:** The nth term of a quadratic sequence can be written as $$a_n = An^2 + Bn + C$$ where $A$, $B$, and $C$ are constants to be determined. 3. **Find the first differences:** Calculate the differences between consecutive terms: $$6 - 1 = 5$$ $$13 - 6 = 7$$ $$22 - 13 = 9$$ $$33 - 22 = 11$$ 4. **Find the second differences:** Calculate the differences between the first differences: $$7 - 5 = 2$$ $$9 - 7 = 2$$ $$11 - 9 = 2$$ Since the second differences are constant and equal to 2, this confirms the sequence is quadratic. 5. **Use the second difference to find $A$:** The second difference equals $2A$, so $$2A = 2 \implies A = 1$$ 6. **Set up equations using the first three terms:** For $n=1$: $$A(1)^2 + B(1) + C = 1 \implies 1 + B + C = 1$$ For $n=2$: $$A(2)^2 + B(2) + C = 6 \implies 4 + 2B + C = 6$$ For $n=3$: $$A(3)^2 + B(3) + C = 13 \implies 9 + 3B + C = 13$$ 7. **Simplify the equations:** From $n=1$: $$1 + B + C = 1 \implies B + C = 0$$ From $n=2$: $$4 + 2B + C = 6 \implies 2B + C = 2$$ From $n=3$: $$9 + 3B + C = 13 \implies 3B + C = 4$$ 8. **Solve the system:** From $B + C = 0$, we get $C = -B$. Substitute into $2B + C = 2$: $$2B - B = 2 \implies B = 2$$ Then, $$C = -B = -2$$ 9. **Write the nth term formula:** $$a_n = n^2 + 2n - 2$$ **Final answer:** The nth term of the sequence is $$\boxed{a_n = n^2 + 2n - 2}$$