Let's start by understanding what is asked in the question. We are given a polynomial [tex]\( P(n) = n^2 + n + 1 \)[/tex] and are asked to find the second differences of the sequence generated by this polynomial.
### Step 1: Generate the Sequence
We first generate the sequence by substituting integer values for [tex]\( n \)[/tex] into the polynomial [tex]\( n^2 + n + 1 \)[/tex]. Let's compute the first few terms:
- For [tex]\( n = 0 \)[/tex]:
[tex]\[
P(0) = 0^2 + 0 + 1 = 1
\][/tex]
- For [tex]\( n = 1 \)[/tex]:
[tex]\[
P(1) = 1^2 + 1 + 1 = 3
\][/tex]
- For [tex]\( n = 2 \)[/tex]:
[tex]\[
P(2) = 2^2 + 2 + 1 = 7
\][/tex]
- For [tex]\( n = 3 \)[/tex]:
[tex]\[
P(3) = 3^2 + 3 + 1 = 13
\][/tex]
- For [tex]\( n = 4 \)[/tex]:
[tex]\[
P(4) = 4^2 + 4 + 1 = 21
\][/tex]
- For [tex]\( n = 5 \)[/tex]:
[tex]\[
P(5) = 5^2 + 5 + 1 = 31
\][/tex]
Thus, the first few terms of the sequence are:
[tex]\[
1, 3, 7, 13, 21, 31
\][/tex]
### Step 2: Calculate the First Differences
The first differences are found by subtracting consecutive terms of the sequence. Let's compute them:
- [tex]\( 3 - 1 = 2 \)[/tex]
- [tex]\( 7 - 3 = 4 \)[/tex]
- [tex]\( 13 - 7 = 6 \)[/tex]
- [tex]\( 21 - 13 = 8 \)[/tex]
- [tex]\( 31 - 21 = 10 \)[/tex]
So, the first differences are:
[tex]\[
2, 4, 6, 8, 10
\][/tex]
### Step 3: Calculate the Second Differences
The second differences are found by subtracting consecutive terms of the first differences. Let's compute them:
- [tex]\( 4 - 2 = 2 \)[/tex]
- [tex]\( 6 - 4 = 2 \)[/tex]
- [tex]\( 8 - 6 = 2 \)[/tex]
- [tex]\( 10 - 8 = 2 \)[/tex]
So, the second differences are:
[tex]\[
2, 2, 2, 2
\][/tex]
Thus, the second differences of the sequence generated by the polynomial [tex]\( n^2 + n + 1 \)[/tex] are:
[tex]\[
2, 2, 2, 2
\][/tex]
