Question

what is the constant value of the second differences for the quadratic polynomial sequence a⌄n = 3n^2 + 2n - 1

Answer 1

To find the constant value of the second differences for the quadratic polynomial sequence given by \( a_n = 3n^2 + 2n - 1 \), we can follow these steps:

1. **Calculate the first differences**:
The first difference \( \Delta a_n \) is defined as:
\[
\Delta a_n = a_{n+1} - a_n
\]
Start by calculating \( a_{n+1} \):
\[
a_{n+1} = 3(n+1)^2 + 2(n+1) - 1 = 3(n^2 + 2n + 1) + 2n + 2 - 1 = 3n^2 + 6n + 3 + 2n + 2 - 1 = 3n^2 + 8n + 4
\]
Now, compute the first difference:
\[
\Delta a_n = a_{n+1} - a_n = (3n^2 + 8n + 4) - (3n^2 + 2n - 1) = 8n + 4 - 2n + 1 = 6n + 5
\]

2. **Calculate the second differences**:
The second difference \( \Delta^2 a_n \) is defined as:
\[
\Delta^2 a_n = \Delta a_{n+1} - \Delta a_n
\]
First, compute \( \Delta a_{n+1} \):
\[
\Delta a_{n+1} = a_{n+2} - a_{n+1}
\]
First, find \( a_{n+2} \):
\[
a_{n+2} = a_{n+1} + \Delta a_{n+1} = a_{n+1} = 3(n+2)^2 + 2(n+2) - 1 = 3(n^2 + 4n + 4) + 2n + 4 - 1 = 3n^2 + 12n + 12 + 2n + 4 - 1 = 3n^2 + 14n + 15
\]
Now calculate \( \Delta a_{n+1} \):
\[
\Delta a_{n+1} = a_{n+2} - a_{n+1} = (3n^2 + 14n + 15) - (3n^2 + 8n + 4) = 14n + 15 - 8n - 4 = 6n + 11
\]
Now compute the second difference:
\[
\Delta^2 a_n = (6n + 11) - (6n + 5) = 6
\]

Thus, the constant value of the second differences for the quadratic polynomial sequence \( a_n = 3n^2 + 2n - 1 \) is **6**.