Question
The sum of a sequence of consecutive numbers from 1 to n is given by the formulia n(n+1)/2 Which level of differences is this polynomial in a constant value?
Answers
GPT-4o mini
The sum of the first \( n \) consecutive numbers is given by the formula:
\[
S(n) = \frac{n(n+1)}{2}
\]
This represents a quadratic polynomial in \( n \). To find the level of differences, we can calculate the differences of the sums for consecutive values of \( n \):
1. **First Differences**: Calculate \( S(n+1) - S(n) \):
\[
S(n+1) = \frac{(n+1)(n+2)}{2}
\]
Now, find \( S(n+1) - S(n) \):
\[
S(n+1) - S(n) = \frac{(n+1)(n+2)}{2} - \frac{n(n+1)}{2} = \frac{(n+1)(n+2 - n)}{2} = \frac{(n+1)(2)}{2} = n + 1
\]
2. **Second Differences**: Now, calculate the first difference of \( n + 1 \):
The first difference of \( n + 1 \) is:
\[
(n + 1) - (n) = 1
\]
Since the second difference is constant (equal to 1), we can conclude:
- The sum \( S(n) \) is a polynomial of degree 2 (quadratic), and the second differences are constant.
Thus, the polynomial \( S(n) = \frac{n(n+1)}{2} \) is a quadratic polynomial, and it has a constant second difference. Therefore, it can be said to be of level 2 in terms of differences, indicating that it is a quadratic polynomial in \( n \).
\[
S(n) = \frac{n(n+1)}{2}
\]
This represents a quadratic polynomial in \( n \). To find the level of differences, we can calculate the differences of the sums for consecutive values of \( n \):
1. **First Differences**: Calculate \( S(n+1) - S(n) \):
\[
S(n+1) = \frac{(n+1)(n+2)}{2}
\]
Now, find \( S(n+1) - S(n) \):
\[
S(n+1) - S(n) = \frac{(n+1)(n+2)}{2} - \frac{n(n+1)}{2} = \frac{(n+1)(n+2 - n)}{2} = \frac{(n+1)(2)}{2} = n + 1
\]
2. **Second Differences**: Now, calculate the first difference of \( n + 1 \):
The first difference of \( n + 1 \) is:
\[
(n + 1) - (n) = 1
\]
Since the second difference is constant (equal to 1), we can conclude:
- The sum \( S(n) \) is a polynomial of degree 2 (quadratic), and the second differences are constant.
Thus, the polynomial \( S(n) = \frac{n(n+1)}{2} \) is a quadratic polynomial, and it has a constant second difference. Therefore, it can be said to be of level 2 in terms of differences, indicating that it is a quadratic polynomial in \( n \).