Question

What are the 2nd differences of the sequence from the following polynomial?

n2+n+1

Answer 1

so what is the answer

Answer 2

Simplifiy what you are trying to say the answer is

Answer 3

To find the second differences of the sequence generated by the polynomial \( n^2 + n + 1 \), we can start by calculating the values of the polynomial for several integer values of \( n \).

Let's calculate \( a_n = n^2 + n + 1 \) for \( n = 0, 1, 2, 3, 4 \):

1. For \( n = 0 \):
\[
a_0 = 0^2 + 0 + 1 = 1
\]

2. For \( n = 1 \):
\[
a_1 = 1^2 + 1 + 1 = 3
\]

3. For \( n = 2 \):
\[
a_2 = 2^2 + 2 + 1 = 7
\]

4. For \( n = 3 \):
\[
a_3 = 3^2 + 3 + 1 = 13
\]

5. For \( n = 4 \):
\[
a_4 = 4^2 + 4 + 1 = 21
\]

Now, the sequence is:
\[
1, 3, 7, 13, 21
\]

Next, we find the first differences \( \Delta a_n \), which are given by \( a_{n+1} - a_n \):

- \( \Delta a_0 = a_1 - a_0 = 3 - 1 = 2 \)
- \( \Delta a_1 = a_2 - a_1 = 7 - 3 = 4 \)
- \( \Delta a_2 = a_3 - a_2 = 13 - 7 = 6 \)
- \( \Delta a_3 = a_4 - a_3 = 21 - 13 = 8 \)

So the first differences are:
\[
2, 4, 6, 8
\]

Next, we find the second differences \( \Delta^2 a_n \), which are given by \( \Delta a_{n+1} - \Delta a_n \):

- \( \Delta^2 a_0 = \Delta a_1 - \Delta a_0 = 4 - 2 = 2 \)
- \( \Delta^2 a_1 = \Delta a_2 - \Delta a_1 = 6 - 4 = 2 \)
- \( \Delta^2 a_2 = \Delta a_3 - \Delta a_2 = 8 - 6 = 2 \)

Thus, the second differences are:
\[
2, 2, 2
\]

The second differences of the sequence given by the polynomial \( n^2 + n + 1 \) are constant and equal to 2.

Answer 4

The second differences of the sequence generated by the polynomial \( n^2 + n + 1 \) are constant and equal to **2**.

Answer 5

The second differences are **2**.