The sum of the first \( n \) consecutive natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
This expression is a polynomial of degree 2 in \( n \).
To investigate the level differences, we can consider the first few values of \( S_n \):
- For \( n = 1 \): \( S_1 = \frac{1(1 + 1)}{2} = 1 \)
- For \( n = 2 \): \( S_2 = \frac{2(2 + 1)}{2} = 3 \)
- For \( n = 3 \): \( S_3 = \frac{3(3 + 1)}{2} = 6 \)
- For \( n = 4 \): \( S_4 = \frac{4(4 + 1)}{2} = 10 \)
- For \( n = 5 \): \( S_5 = \frac{5(5 + 1)}{2} = 15 \)
Now let's calculate the first differences:
- \( S_2 - S_1 = 3 - 1 = 2 \)
- \( S_3 - S_2 = 6 - 3 = 3 \)
- \( S_4 - S_3 = 10 - 6 = 4 \)
- \( S_5 - S_4 = 15 - 10 = 5 \)
First differences: \( 2, 3, 4, 5 \)
Next, we calculate the second differences:
- \( 3 - 2 = 1 \)
- \( 4 - 3 = 1 \)
- \( 5 - 4 = 1 \)
Second differences: \( 1, 1, 1 \)
We see that the second differences are constant, which indicates that the original polynomial \( S_n = \frac{n(n + 1)}{2} \) is a quadratic polynomial.
So to answer the question, the level differences become constant at the second level.
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