Quadratic Sequence 43Df1C

1. **State the problem:** We are given the first 6 terms of a quadratic sequence: $-3, 3, 13, 27, 45, 67$. We need to find a formula for the $n$th term, $T_n$. 2. **Recall the formula for a quadratic sequence:** The $n$th term of a quadratic sequence can be written as: $$T_n = an^2 + bn + c$$ where $a$, $b$, and $c$ are constants to be determined. 3. **Find the first differences:** $3 - (-3) = 6$ $13 - 3 = 10$ $27 - 13 = 14$ $45 - 27 = 18$ $67 - 45 = 22$ So the first differences are: $6, 10, 14, 18, 22$. 4. **Find the second differences:** $10 - 6 = 4$ $14 - 10 = 4$ $18 - 14 = 4$ $22 - 18 = 4$ The second differences are constant and equal to $4$. 5. **Use the second difference to find $a$:** The second difference of a quadratic sequence is always $2a$. So, $$2a = 4 \implies a = 2$$ 6. **Set up equations to find $b$ and $c$ using the first two terms:** For $n=1$, $$T_1 = a(1)^2 + b(1) + c = 2 + b + c = -3$$ For $n=2$, $$T_2 = a(2)^2 + b(2) + c = 8 + 2b + c = 3$$ 7. **Write the system of equations:** $$2 + b + c = -3$$ $$8 + 2b + c = 3$$ 8. **Simplify the system:** From the first equation: $$b + c = -5$$ From the second equation: $$2b + c = -5$$ 9. **Subtract the first from the second:** $$(2b + c) - (b + c) = -5 - (-5)$$ $$b = 0$$ 10. **Find $c$ using $b=0$ in $b + c = -5$:** $$0 + c = -5 \implies c = -5$$ 11. **Write the final formula:** $$T_n = 2n^2 + 0 \cdot n - 5 = 2n^2 - 5$$ **Answer:** The $n$th term of the sequence is $$\boxed{2n^2 - 5}$$