Question

Find the 2nd differences of the sequence represented by the polynomial.
an = 4n^2 + 3n - 7

A. 0
B. 8
C. 15
D. 4

Answer 1

To find the 2nd differences of the sequence defined by the polynomial \( a_n = 4n^2 + 3n - 7 \), we first need to calculate the first differences and then the second differences.

1. **Calculate the values of \( a_n \):**
- For \( n = 1 \):
\[
a_1 = 4(1)^2 + 3(1) - 7 = 4 + 3 - 7 = 0
\]
- For \( n = 2 \):
\[
a_2 = 4(2)^2 + 3(2) - 7 = 16 + 6 - 7 = 15
\]
- For \( n = 3 \):
\[
a_3 = 4(3)^2 + 3(3) - 7 = 36 + 9 - 7 = 38
\]
- For \( n = 4 \):
\[
a_4 = 4(4)^2 + 3(4) - 7 = 64 + 12 - 7 = 69
\]

The values of the sequence are:
\[
a_1 = 0, \quad a_2 = 15, \quad a_3 = 38, \quad a_4 = 69
\]

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

The first differences are:
\[
d_1 = 15, \quad d_2 = 23, \quad d_3 = 31
\]

3. **Calculate the second differences:**
- \( d_2 - d_1 = 23 - 15 = 8 \)
- \( d_3 - d_2 = 31 - 23 = 8 \)

The second differences are:
\[
e_1 = 8, \quad e_2 = 8
\]

Since the second differences are constant:

The answer is:
\[
\boxed{8}
\]