Question

the sum of a sequence of consecutive numbers from 1 to n is given by the formula n(n+1)/2. Which level differences of this polynomial is a constant value?

Answer 1

To find the level differences of the polynomial representing the sum of the first \( n \) consecutive integers, given by

\[
S(n) = \frac{n(n+1)}{2},
\]

we first note that this is a quadratic polynomial in \( n \).

Let's rewrite the polynomial:

\[
S(n) = \frac{1}{2}n^2 + \frac{1}{2}n.
\]

Next, we can compute the first few level differences (also known as finite differences):

1. **First Level Differences**: This is the difference between consecutive values of \( S(n) \):

\[
S(n+1) - S(n).
\]

Calculating this:

\[
S(n+1) = \frac{(n+1)((n+1)+1)}{2} = \frac{(n+1)(n+2)}{2},
\]

\[
S(n) = \frac{n(n+1)}{2}.
\]

Finding the difference:

\[
S(n+1) - S(n) = \frac{(n+1)(n+2)}{2} - \frac{n(n+1)}{2}.
\]

Simplifying:

\[
S(n+1) - S(n) = \frac{(n+1)(n+2) - n(n+1)}{2} = \frac{(n^2 + 3n + 2 - n^2 - n)}{2} = \frac{2n + 2}{2} = n + 1.
\]

2. **Second Level Differences**: Now, we find the second level differences by taking the differences of the first level differences:

\[
(n+1) - n = 1.
\]

Since the second level differences are constant (equal to 1), we conclude that the second level differences of the polynomial \( S(n) \) are constant.

In summary, the level differences of the polynomial \( S(n) \) are constant after the **second level differences**. Thus, the answer is:

The second level differences of the polynomial are constant.