2

Question

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!