B1 Calculate: $1000^{2}-999^{2}+998^{2}-997^{2}+\ldots+4^{2}-3^{2}+2^{2}-1^{2}$
B1. $\quad a^{2}-b^{2}=(a-b)(a+b)$ so $1000^{2}-999^{2}=1 \times 1999$. The desired sum is thus equal to: $1999+1995+1991+\ldots+7+3$. $\quad$ These are 500 numbers $(2000,1996, \ldots, 4$ or $500,499, \ldots, 1)$. Sum $=1999+1995+\ldots+7+3$

Sum $=3+7+\ldots+1995+1999$ Add vertically; each pair gives 2002.

$2 \times$ Sum $=2002+2002+\ldots+2002+2002$

For the Sum, we find 500 times 2002 divided by 2, so $500 \times 1001=500500$.