Question

2
8
18
32
50
6
10
14
18
Write an expression in standard form that represents the sequence.
(1 point)
---x^2+--x+---

Answer 1

To find the expression in standard form that represents the given sequence, we can analyze the sequence as follows:

The first part of the sequence is:
- 2, 8, 18, 32, 50

The second part of the sequence is:
- 6, 10, 14, 18

Let's analyze the first part:

1. **Identifying the first part:**
- The differences between the consecutive terms are:
- 8 - 2 = 6
- 18 - 8 = 10
- 32 - 18 = 14
- 50 - 32 = 18

The first differences are:
- 6, 10, 14, 18

2. **Identifying the second part (differences of the first differences):**
- 10 - 6 = 4
- 14 - 10 = 4
- 18 - 14 = 4

The second differences are constant (4), indicating that the original sequence can be represented by a quadratic polynomial of the form \( ax^2 + bx + c \).

3. **Finding the coefficients a, b, and c:**
- The second difference (4) is equal to \( 2a \):
- Thus, \( 2a = 4 \) implies \( a = 2 \).

4. **Next, we can use the known values to find b and c. Let's write the quadratic in the form:**
\[
f(n) = 2n^2 + bn + c
\]

We can use the known values of the sequence to set up equations.

5. **Using the first few values:**
- \( f(1) = 2 \) implies \( 2(1)^2 + b(1) + c = 2 \) → \( 2 + b + c = 2 \) → \( b + c = 0 \) (Equation 1)
- \( f(2) = 8 \) implies \( 2(2)^2 + b(2) + c = 8 \) → \( 8 + 2b + c = 8 \) → \( 2b + c = 0 \) (Equation 2)
- \( f(3) = 18 \) implies \( 2(3)^2 + b(3) + c = 18 \) → \( 18 + 3b + c = 18 \) → \( 3b + c = 0 \) (Equation 3)

6. **Solving the equations:**
From **Equation 1**, we have \( c = -b \).

Substitute \( c = -b \) into **Equation 2**:
\[
2b - b = 0 \implies b = 0
\]

Substitute \( b = 0 \) back into **Equation 1**:
\[
0 + c = 0 \implies c = 0
\]

Therefore, the coefficients are:
- \( a = 2 \)
- \( b = 0 \)
- \( c = 0 \)

Thus, the expression representing the sequence is:
\[
\boxed{2x^2}
\]

If you need help with the second sequence (6, 10, 14, 18), let me know!