Question
At which differences does the following polynomial sequence reach a constant value? an=2n4−n3
Answers
GPT-4o mini
To determine at which differences the polynomial sequence given by \( a_n = 2n^4 - n^3 \) reaches a constant value, we need to compute the differences of the sequence.
We start with the first differences, which are calculated as follows:
\[
\Delta a_n = a_{n+1} - a_n
\]
Calculating \( a_{n+1} \):
\[
a_{n+1} = 2(n+1)^4 - (n+1)^3
\]
Expanding this:
\[
(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1
\]
\[
(n+1)^3 = n^3 + 3n^2 + 3n + 1
\]
Thus, we have:
\[
a_{n+1} = 2(n^4 + 4n^3 + 6n^2 + 4n + 1) - (n^3 + 3n^2 + 3n + 1)
\]
\[
= 2n^4 + 8n^3 + 12n^2 + 8n + 2 - n^3 - 3n^2 - 3n - 1
\]
\[
= 2n^4 + (8n^3 - n^3) + (12n^2 - 3n^2) + (8n - 3n) + (2 - 1)
\]
\[
= 2n^4 + 7n^3 + 9n^2 + 5n + 1
\]
Now, we compute the first difference:
\[
\Delta a_n = a_{n+1} - a_n = (2n^4 + 7n^3 + 9n^2 + 5n + 1) - (2n^4 - n^3)
\]
\[
= 7n^3 + 9n^2 + 5n + 1 + n^3
\]
\[
= 8n^3 + 9n^2 + 5n + 1
\]
Next, we compute the second differences:
\[
\Delta^2 a_n = \Delta a_{n+1} - \Delta a_n
\]
Calculating \( \Delta a_{n+1} \):
\[
\Delta a_{n+1} = a_{n+2} - a_{n+1}
\]
We follow a similar process as above to find \( a_{n+2} \) and then find \( \Delta a_{n+1} \).
But let's recognize that if \( a_n \) is a polynomial of degree 4, then \( \Delta a_n \) will be a polynomial of degree 3, \( \Delta^2 a_n \) will be a polynomial of degree 2, and \( \Delta^3 a_n \) will be a polynomial of degree 1. Finally, \( \Delta^4 a_n \) will give us a constant.
To summarize:
- \( a_n = 2n^4 - n^3 \) is a polynomial of degree 4.
- The first differences \( \Delta a_n \) yield a polynomial of degree 3.
- The second differences \( \Delta^2 a_n \) yield a polynomial of degree 2.
- The third differences \( \Delta^3 a_n \) yield a polynomial of degree 1.
- The fourth differences \( \Delta^4 a_n \) yield a constant value.
Thus, the polynomial sequence \( a_n = 2n^4 - n^3 \) reaches a constant value at the fourth differences.
We start with the first differences, which are calculated as follows:
\[
\Delta a_n = a_{n+1} - a_n
\]
Calculating \( a_{n+1} \):
\[
a_{n+1} = 2(n+1)^4 - (n+1)^3
\]
Expanding this:
\[
(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1
\]
\[
(n+1)^3 = n^3 + 3n^2 + 3n + 1
\]
Thus, we have:
\[
a_{n+1} = 2(n^4 + 4n^3 + 6n^2 + 4n + 1) - (n^3 + 3n^2 + 3n + 1)
\]
\[
= 2n^4 + 8n^3 + 12n^2 + 8n + 2 - n^3 - 3n^2 - 3n - 1
\]
\[
= 2n^4 + (8n^3 - n^3) + (12n^2 - 3n^2) + (8n - 3n) + (2 - 1)
\]
\[
= 2n^4 + 7n^3 + 9n^2 + 5n + 1
\]
Now, we compute the first difference:
\[
\Delta a_n = a_{n+1} - a_n = (2n^4 + 7n^3 + 9n^2 + 5n + 1) - (2n^4 - n^3)
\]
\[
= 7n^3 + 9n^2 + 5n + 1 + n^3
\]
\[
= 8n^3 + 9n^2 + 5n + 1
\]
Next, we compute the second differences:
\[
\Delta^2 a_n = \Delta a_{n+1} - \Delta a_n
\]
Calculating \( \Delta a_{n+1} \):
\[
\Delta a_{n+1} = a_{n+2} - a_{n+1}
\]
We follow a similar process as above to find \( a_{n+2} \) and then find \( \Delta a_{n+1} \).
But let's recognize that if \( a_n \) is a polynomial of degree 4, then \( \Delta a_n \) will be a polynomial of degree 3, \( \Delta^2 a_n \) will be a polynomial of degree 2, and \( \Delta^3 a_n \) will be a polynomial of degree 1. Finally, \( \Delta^4 a_n \) will give us a constant.
To summarize:
- \( a_n = 2n^4 - n^3 \) is a polynomial of degree 4.
- The first differences \( \Delta a_n \) yield a polynomial of degree 3.
- The second differences \( \Delta^2 a_n \) yield a polynomial of degree 2.
- The third differences \( \Delta^3 a_n \) yield a polynomial of degree 1.
- The fourth differences \( \Delta^4 a_n \) yield a constant value.
Thus, the polynomial sequence \( a_n = 2n^4 - n^3 \) reaches a constant value at the fourth differences.