Question

Using suitable algebraic identities, find the value of 2004^2 - 2003^2 + 2002^2 - 2001^2

Answer 1

$(x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1$

Let $x=2003$. Then

$2004^2-2003^2=(2003+1)^2-2003^2=2\cdot2003+1=4007$

Similarly, if $x=2001$, then

$2002^2-2001^2=2\cdot2001+1=4003$

So

$2004^2-2003^2+2002^2-2001^2=4007+4003=8010$

Alternatively, you can consider the larger expression

$(x+3)^2-(x+2)^2+(x+1)^2-x^2$

Expanding each binomial product gives

$(x^2+6x+9)-(x^2+4x+4)+(x^2+2x+1)-x^2=4x+6$

Then if $x=2001$, we get

$2004^2-2003^2+2002^2-2001^2=4\cdot2001+6=8010$