Question

Find the 2nd differences of the sequence represented by the polynomial. an=4n2+3n−7 (1 point) Responses 0 0 8 8 15 15 4 4

Answer 1

To find the second differences of the sequence represented by the polynomial \( a_n = 4n^2 + 3n - 7 \), we can follow these steps:

1. **Calculate the first few terms of the sequence**:
- For \( n = 1 \):
\[
a_1 = 4(1)^2 + 3(1) - 7 = 4 + 3 - 7 = 0
\]
- For \( n = 2 \):
\[
a_2 = 4(2)^2 + 3(2) - 7 = 4(4) + 6 - 7 = 16 + 6 - 7 = 15
\]
- For \( n = 3 \):
\[
a_3 = 4(3)^2 + 3(3) - 7 = 4(9) + 9 - 7 = 36 + 9 - 7 = 38
\]
- For \( n = 4 \):
\[
a_4 = 4(4)^2 + 3(4) - 7 = 4(16) + 12 - 7 = 64 + 12 - 7 = 69
\]
- For \( n = 5 \):
\[
a_5 = 4(5)^2 + 3(5) - 7 = 4(25) + 15 - 7 = 100 + 15 - 7 = 108
\]

So the first few terms of the sequence are:
- \( a_1 = 0 \)
- \( a_2 = 15 \)
- \( a_3 = 38 \)
- \( a_4 = 69 \)
- \( a_5 = 108 \)

2. **Calculate the first differences**:
- \( a_2 - a_1 = 15 - 0 = 15 \)
- \( a_3 - a_2 = 38 - 15 = 23 \)
- \( a_4 - a_3 = 69 - 38 = 31 \)
- \( a_5 - a_4 = 108 - 69 = 39 \)

The first differences are:
- \( 15, 23, 31, 39 \)

3. **Calculate the second differences**:
- \( 23 - 15 = 8 \)
- \( 31 - 23 = 8 \)
- \( 39 - 31 = 8 \)

The second differences are:
- \( 8, 8, 8 \)

Since the second differences are constant and equal to \( 8 \), they confirm that the original sequence is quadratic.

The answer is **8**.