Sequences - Minimally Different - Free Printable

To find the \( n \)-th term of a quadratic sequence, we need to determine the general formula for the sequence. A quadratic sequence is one where the second differences between consecutive terms are constant. The general form of the \( n \)-th term of a quadratic sequence is:

\[
a_n = an^2 + bn + c
\]

where \( a \), \( b \), and \( c \) are constants to be determined.

Step-by-Step Solution for Each Sequence

#### 1. Sequence: 2, 8, 18, 32, 50, ...

1. First Differences:
\[
8 - 2 = 6, \quad 18 - 8 = 10, \quad 32 - 18 = 14, \quad 50 - 32 = 18
\]
First differences: \( 6, 10, 14, 18 \)

2. Second Differences:
\[
10 - 6 = 4, \quad 14 - 10 = 4, \quad 18 - 14 = 4
\]
Second differences: \( 4, 4, 4 \) (constant)

3. General Form:
Since the second differences are constant, the sequence is quadratic. Assume:
\[
a_n = an^2 + bn + c
\]

4. Set up equations using the first few terms:
- For \( n = 1 \): \( a(1)^2 + b(1) + c = 2 \)
\[
a + b + c = 2 \quad \text{(Equation 1)}
\]
- For \( n = 2 \): \( a(2)^2 + b(2) + c = 8 \)
\[
4a + 2b + c = 8 \quad \text{(Equation 2)}
\]
- For \( n = 3 \): \( a(3)^2 + b(3) + c = 18 \)
\[
9a + 3b + c = 18 \quad \text{(Equation 3)}
\]

5. Solve the system of equations:
- Subtract Equation 1 from Equation 2:
\[
(4a + 2b + c) - (a + b + c) = 8 - 2
\]
\[
3a + b = 6 \quad \text{(Equation 4)}
\]
- Subtract Equation 2 from Equation 3:
\[
(9a + 3b + c) - (4a + 2b + c) = 18 - 8
\]
\[
5a + b = 10 \quad \text{(Equation 5)}
\]

6. Solve Equations 4 and 5:
- Subtract Equation 4 from Equation 5:
\[
(5a + b) - (3a + b) = 10 - 6
\]
\[
2a = 4 \implies a = 2
\]
- Substitute \( a = 2 \) into Equation 4:
\[
3(2) + b = 6
\]
\[
6 + b = 6 \implies b = 0
\]
- Substitute \( a = 2 \) and \( b = 0 \) into Equation 1:
\[
2 + 0 + c = 2 \implies c = 0
\]

7. Final Formula:
\[
a_n = 2n^2
\]

#### 2. Sequence: 3, 9, 19, 33, 51, ...

1. First Differences:
\[
9 - 3 = 6, \quad 19 - 9 = 10, \quad 33 - 19 = 14, \quad 51 - 33 = 18
\]
First differences: \( 6, 10, 14, 18 \)

2. Second Differences:
\[
10 - 6 = 4, \quad 14 - 10 = 4, \quad 18 - 14 = 4
\]
Second differences: \( 4, 4, 4 \) (constant)

3. General Form:
\[
a_n = an^2 + bn + c
\]

4. Set up equations using the first few terms:
- For \( n = 1 \): \( a(1)^2 + b(1) + c = 3 \)
\[
a + b + c = 3 \quad \text{(Equation 1)}
\]
- For \( n = 2 \): \( a(2)^2 + b(2) + c = 9 \)
\[
4a + 2b + c = 9 \quad \text{(Equation 2)}
\]
- For \( n = 3 \): \( a(3)^2 + b(3) + c = 19 \)
\[
9a + 3b + c = 19 \quad \text{(Equation 3)}
\]

5. Solve the system of equations:
- Subtract Equation 1 from Equation 2:
\[
(4a + 2b + c) - (a + b + c) = 9 - 3
\]
\[
3a + b = 6 \quad \text{(Equation 4)}
\]
- Subtract Equation 2 from Equation 3:
\[
(9a + 3b + c) - (4a + 2b + c) = 19 - 9
\]
\[
5a + b = 10 \quad \text{(Equation 5)}
\]

6. Solve Equations 4 and 5:
- Subtract Equation 4 from Equation 5:
\[
(5a + b) - (3a + b) = 10 - 6
\]
\[
2a = 4 \implies a = 2
\]
- Substitute \( a = 2 \) into Equation 4:
\[
3(2) + b = 6
\]
\[
6 + b = 6 \implies b = 0
\]
- Substitute \( a = 2 \) and \( b = 0 \) into Equation 1:
\[
2 + 0 + c = 3 \implies c = 1
\]

7. Final Formula:
\[
a_n = 2n^2 + 1
\]

#### 3. Sequence: 4, 11, 22, 37, 56, ...

1. First Differences:
\[
11 - 4 = 7, \quad 22 - 11 = 11, \quad 37 - 22 = 15, \quad 56 - 37 = 19
\]
First differences: \( 7, 11, 15, 19 \)

2. Second Differences:
\[
11 - 7 = 4, \quad 15 - 11 = 4, \quad 19 - 15 = 4
\]
Second differences: \( 4, 4, 4 \) (constant)

3. General Form:
\[
a_n = an^2 + bn + c
\]

4. Set up equations using the first few terms:
- For \( n = 1 \): \( a(1)^2 + b(1) + c = 4 \)
\[
a + b + c = 4 \quad \text{(Equation 1)}
\]
- For \( n = 2 \): \( a(2)^2 + b(2) + c = 11 \)
\[
4a + 2b + c = 11 \quad \text{(Equation 2)}
\]
- For \( n = 3 \): \( a(3)^2 + b(3) + c = 22 \)
\[
9a + 3b + c = 22 \quad \text{(Equation 3)}
\]

5. Solve the system of equations:
- Subtract Equation 1 from Equation 2:
\[
(4a + 2b + c) - (a + b + c) = 11 - 4
\]
\[
3a + b = 7 \quad \text{(Equation 4)}
\]
- Subtract Equation 2 from Equation 3:
\[
(9a + 3b + c) - (4a + 2b + c) = 22 - 11
\]
\[
5a + b = 11 \quad \text{(Equation 5)}
\]

6. Solve Equations 4 and 5:
- Subtract Equation 4 from Equation 5:
\[
(5a + b) - (3a + b) = 11 - 7
\]
\[
2a = 4 \implies a = 2
\]
- Substitute \( a = 2 \) into Equation 4:
\[
3(2) + b = 7
\]
\[
6 + b = 7 \implies b = 1
\]
- Substitute \( a = 2 \) and \( b = 1 \) into Equation 1:
\[
2 + 1 + c = 4 \implies c = 1
\]

7. Final Formula:
\[
a_n = 2n^2 + n + 1
\]

Final Answer:

\[
\boxed{2n^2}
\]