The given problem involves calculating the sum of a sequence:
$100^2 - 99^2 + 98^2 - 97^2 + \dots + 2^2 - 1.$

Step 1: Simplify Each Pair

The pattern alternates between squares being subtracted and added. Each pair $(a^2 - b^2)$ can be simplified using the difference of squares formula: $a^2 - b^2 = (a + b)(a - b).$

For the given problem: $100^2 - 99^2 = (100 + 99)(100 - 99) = 199 \cdot 1 = 199,$ $98^2 - 97^2 = (98 + 97)(98 - 97) = 195 \cdot 1 = 195,$ and so on.

Step 2: Recognize the Pattern

This sequence continues down to: $2^2 - 1^2 = (2 + 1)(2 - 1) = 3 \cdot 1 = 3.$

The terms of the sequence are: $199, 195, 191, \dots, 3.$

Step 3: Find the Number of Terms

This is an arithmetic sequence where the first term is $199$, the common difference is $-4$, and the last term is $3$. Let $n$ be the number of terms. Using the formula for the $n$-th term of an arithmetic sequence: $a_n = a_1 + (n - 1)d,$ where $a_n = 3$, $a_1 = 199$, and $d = -4$: $3 = 199 + (n - 1)(-4),$ $3 = 199 - 4n + 4,$ $3 = 203 - 4n,$ $4n = 203 - 3 = 200,$ $n = 50.$

Step 4: Sum the Arithmetic Sequence

The sum of an arithmetic sequence is given by: $S_n = \frac{n}{2}(a_1 + a_n),$ where $n = 50$, $a_1 = 199$, and $a_n = 3$: $S_{50} = \frac{50}{2}(199 + 3) = 25 \cdot 202 = 5050.$

Final Answer:

The sum is $\boxed{5050}$.

Do you have any questions or need further explanation?

Here are 5 related questions to practice:

1. How would you find the sum of $100^2 - 99^2 + 98^2 - 97^2 + \dots + 4^2 - 3^2$?
2. Can you prove the difference of squares formula?
3. What is the sum of the first 50 even numbers?
4. How does the sum of an arithmetic sequence change when the common difference is positive?
5. Can you derive the formula for the sum of a geometric sequence?

Tip: For alternating sums, always look for patterns or formulas like the difference of squares to simplify computations.