Sure! Let's find the second differences for the sequence generated by the polynomial [tex]\( n^2 + n + 1 \)[/tex].
### Step-by-Step Solution:
#### Step 1: Generate the sequence
The first step is to generate the sequence by substituting integer values of [tex]\( n \)[/tex] into the polynomial. Let's evaluate the polynomial for [tex]\( n = 0, 1, 2, 3, 4, \)[/tex] and 5:
For [tex]\( n = 0 \)[/tex]:
[tex]\[ n^2 + n + 1 = 0^2 + 0 + 1 = 1 \][/tex]
For [tex]\( n = 1 \)[/tex]:
[tex]\[ n^2 + n + 1 = 1^2 + 1 + 1 = 3 \][/tex]
For [tex]\( n = 2 \)[/tex]:
[tex]\[ n^2 + n + 1 = 2^2 + 2 + 1 = 7 \][/tex]
For [tex]\( n = 3 \)[/tex]:
[tex]\[ n^2 + n + 1 = 3^2 + 3 + 1 = 13 \][/tex]
For [tex]\( n = 4 \)[/tex]:
[tex]\[ n^2 + n + 1 = 4^2 + 4 + 1 = 21 \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ n^2 + n + 1 = 5^2 + 5 + 1 = 31 \][/tex]
Thus, the sequence generated is:
[tex]\[ 1, 3, 7, 13, 21, 31 \][/tex]
#### Step 2: Calculate the first differences
To compute the first differences, subtract each term in the sequence from the term that follows it:
[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
To find the second differences, we subtract each term in the first differences from the term that follows it:
[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]
### Conclusion
Therefore, the second differences of the sequence generated by the polynomial [tex]\( n^2 + n + 1 \)[/tex] are:
[tex]\[ 2, 2, 2, 2 \][/tex]
