Question

Calcule $2002^2-2001^2+2000^2-1999^2+\dots+2^2-1^2$.

Answer 1

$a^2-b^2=(a+b)(a-b)$

$2~002^2-2~001^2=(2~002+2~001)(2~002-2~001)=4~003$

$2~000^2-1~999^2=(2~000+1~999)(2~000-1~999)=3~999$

$1~998^2-1~997^2=(1~998+1~997)(1~998-1~997)=3~995$

$1~996^2-1~995^2=(1~996+1~995)(1~996-1~995)=3~991$

$\dots$

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

$1~001$ pares.

$S=\underbrace{3+7+\dots+3~991+3~995+3~999+4~003}_{1001~parcelas}}$

$S=\dfrac{(3+4~003)1~001}{2}=\dfarc{4~006\cdot1~001}{2}=2~005~003$