To find the $n$-th term of the quadratic sequence $7, 10, 15, 22, 31, \dots$, we follow these steps:

Step 1: Identify the differences

• First differences: $10 - 7 = 3, \; 15 - 10 = 5, \; 22 - 15 = 7, \; 31 - 22 = 9$ So, the first differences are: $3, 5, 7, 9$.

• Second differences: $5 - 3 = 2, \; 7 - 5 = 2, \; 9 - 7 = 2$ Since the second differences are constant, the sequence is quadratic.

Step 2: General form of a quadratic sequence

The general form of a quadratic sequence is: $a_n = An^2 + Bn + C$ We need to find $A$, $B$, and $C$.

Step 3: Form equations using the sequence

Substitute the terms of the sequence into the general form:

• For $n = 1$, $a_1 = 7$: $A(1)^2 + B(1) + C = 7 \quad \Rightarrow \quad A + B + C = 7$
• For $n = 2$, $a_2 = 10$: $A(2)^2 + B(2) + C = 10 \quad \Rightarrow \quad 4A + 2B + C = 10$
• For $n = 3$, $a_3 = 15$: $A(3)^2 + B(3) + C = 15 \quad \Rightarrow \quad 9A + 3B + C = 15$

Step 4: Solve the system of equations

1. From $A + B + C = 7$, $C = 7 - A - B$
2. Substitute $C = 7 - A - B$ into $4A + 2B + C = 10$: $4A + 2B + (7 - A - B) = 10$ Simplify: $3A + B + 7 = 10 \quad \Rightarrow \quad 3A + B = 3$ $B = 3 - 3A$
3. Substitute $C = 7 - A - B$ and $B = 3 - 3A$ into $9A + 3B + C = 15$: $9A + 3(3 - 3A) + (7 - A - (3 - 3A)) = 15$ Simplify: $9A + 9 - 9A + 7 - A - 3 + 3A = 15$ $A + 13 = 15 \quad \Rightarrow \quad A = 2$
4. Find $B$ and $C$: $B = 3 - 3(2) = -3, \quad C = 7 - 2 - (-3) = 8$

Step 5: Write the $n$-th term

The $n$-th term is: $a_n = 2n^2 - 3n + 8$

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Follow-Up Questions:

1. Can you derive the formula for any quadratic sequence?
2. What changes in the process if the second differences are not constant?
3. How can you verify the $n$-th term formula with the given sequence?
4. How do linear and quadratic sequences differ in terms of differences?
5. Can a cubic sequence be solved similarly?

Tip:

Always double-check your coefficients by substituting them back into the formula to ensure all terms match the given sequence.