Count

100² - 99² + 98² - 97² + ... + 2² - 1²

=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+98+97+...+2+1

=5050

We know that the difference of the squares of two consecutive natural numbers is always equal to the sum of the numbers.

PROOF

Let the consecutive natural numbers be $n$ and $(n+1)$ respectively, then we have the difference of their squares as:

$(n+1)^{2} - n^2 = (n+1+n)(n+1-n) = 2n+1 = n+(n+1)$ , which is clearly the sum of the two numbers.

Using this proof, we try to find out the value of the expression by grouping its components like this:

$(100^{2}-99^{2})+(98^{2}-97^{2})+...(2^{2}-1^{2})$

Clearly, the groups are the differences of two consecutive positive integers. Applying our proof, the above expression advances to:

$=(100+99)+(98+97)+...(2+1)=100+99+98+...+3+2+1$

Thanks to Gauss! We know from his hilarious attempt of finding the sum of the first 100 natural numbers, which equals $\boxed{5050}$ . We can still find the sum of the first 100 natural numbers by finding $S_{n}$ where first term $a=1$ , no. of terms $n=100$ and last term $l=a_{n}=100$ .

Great problem! Richard said it all! :D