To find the standard form polynomial that represents the given sequence, we can use the information from the successive differences.
The original sequence is:
- 6
- 23
- 46
- 75
- 110
The first differences are:
- 23 - 6 = 17
- 46 - 23 = 23
- 75 - 46 = 29
- 110 - 75 = 35
The second differences are:
- 23 - 17 = 6
- 29 - 23 = 6
- 35 - 29 = 6
Since the second differences are constant (equal to 6), this indicates that the original sequence can be represented by a quadratic polynomial of the form \( ax^2 + bx + c \).
We start with the second difference:
- The second difference being constant and equal to \( 6 \) means \( 2a = 6 \).
- So, \( a = 3 \).
Next, we use the information from the sequence to find coefficients \( b \) and \( c \). We know three points from the sequence:
- When \( x = 1 \), \( f(1) = 6 \)
- When \( x = 2 \), \( f(2) = 23 \)
- When \( x = 3 \), \( f(3) = 46 \)
Using the polynomial form \( f(x) = 3x^2 + bx + c \):
-
For \( x = 1 \): \[ 3(1)^2 + b(1) + c = 6 \] \[ 3 + b + c = 6 \quad \text{(1)} \] \[ b + c = 3 \quad \text{(simplified)} \]
-
For \( x = 2 \): \[ 3(2)^2 + b(2) + c = 23 \] \[ 3(4) + 2b + c = 23 \] \[ 12 + 2b + c = 23 \] \[ 2b + c = 11 \quad \text{(2)} \]
-
For \( x = 3 \): \[ 3(3)^2 + b(3) + c = 46 \] \[ 3(9) + 3b + c = 46 \] \[ 27 + 3b + c = 46 \] \[ 3b + c = 19 \quad \text{(3)} \]
Now we can solve the system of equations made from equations (1), (2), and (3).
From (1): \[ c = 3 - b \]
Substituting \( c \) into (2): \[ 2b + (3 - b) = 11 \] \[ 2b + 3 - b = 11 \] \[ b + 3 = 11 \] \[ b = 8 \]
Now substituting \( b = 8 \) back into (1): \[ 8 + c = 3 \] \[ c = 3 - 8 = -5 \]
Thus, we have found:
- \( a = 3 \)
- \( b = 8 \)
- \( c = -5 \)
The polynomial that represents the sequence is: \[ f(x) = 3x^2 + 8x - 5 \]
Therefore, the standard form polynomial that represents the sequence is:
C. 3x² + 8x - 5.
1 answer