[tex](x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1[/tex]
Let [tex]x=2003[/tex]. Then
[tex]2004^2-2003^2=(2003+1)^2-2003^2=2\cdot2003+1=4007[/tex]
Similarly, if [tex]x=2001[/tex], then
[tex]2002^2-2001^2=2\cdot2001+1=4003[/tex]
So
[tex]2004^2-2003^2+2002^2-2001^2=4007+4003=8010[/tex]
Alternatively, you can consider the larger expression
[tex](x+3)^2-(x+2)^2+(x+1)^2-x^2[/tex]
Expanding each binomial product gives
[tex](x^2+6x+9)-(x^2+4x+4)+(x^2+2x+1)-x^2=4x+6[/tex]
Then if [tex]x=2001[/tex], we get
[tex]2004^2-2003^2+2002^2-2001^2=4\cdot2001+6=8010[/tex]
