Certainly! Let's solve the problem step-by-step.
1. Define the polynomial function and generate the sequence:
The polynomial given is [tex]\( n^2 + n + 1 \)[/tex]. We will compute this polynomial for a few values of [tex]\( n \)[/tex]. Let's select [tex]\( n \)[/tex] values from 1 to 5 for simplicity.
For [tex]\( n = 1 \)[/tex]:
[tex]\[
1^2 + 1 + 1 = 3
\][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[
2^2 + 2 + 1 = 7
\][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[
3^2 + 3 + 1 = 13
\][/tex]
For [tex]\( n = 4 \)[/tex]:
[tex]\[
4^2 + 4 + 1 = 21
\][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[
5^2 + 5 + 1 = 31
\][/tex]
So, the sequence is:
[tex]\[
[3, 7, 13, 21, 31]
\][/tex]
2. Calculate the first differences:
The first differences are obtained by subtracting consecutive terms of the sequence.
[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]\[
[4, 6, 8, 10]
\][/tex]
3. Calculate the second differences:
The second differences are obtained by subtracting consecutive first differences.
[tex]\[
6 - 4 = 2
\][/tex]
[tex]\[
8 - 6 = 2
\][/tex]
[tex]\[
10 - 8 = 2
\][/tex]
So, the second differences are:
[tex]\[
[2, 2, 2]
\][/tex]
Therefore, the second differences of the sequence generated from the polynomial [tex]\( n^2 + n + 1 \)[/tex] are [tex]\([2, 2, 2]\)[/tex].
