To solve the problem of finding the difference between the sum of the first 100 even numbers and the sum of the first 100 odd numbers, we can use the following reasoning:
Sum of first (n) even numbers: The first (n) even numbers are (2, 4, 6, ..., 2n). This is an arithmetic sequence with first term (a=2), last term (l=2n), and number of terms (n). The sum is:
[
S_{\text{even}} = \frac{n}{2}(a + l) = \frac{n}{2}(2 + 2n) = n(n + 1)
]
Sum of first (n) odd numbers: The first (n) odd numbers are (1, 3, 5, ..., 2n-1). This is also an arithmetic sequence with first term (a=1), last term (l=2n-1), and number of terms (n). The sum is:
[
S_{\text{odd}} = \frac{n}{2}(1 + (2n-1)) = \frac{n}{2}(2n) = n^2
]
[ \text{Difference} = S{\text{even}} - S{\text{odd}} = n(n + 1) - n^2 = n^2 + n - n^2 = n ]
For (n=100), the difference is (100).
Answer: (\boxed{100})
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