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`