Question

Tính nhanh: 1^2-2^2+3^2-4^2+....-2004^2+2005^2

Answer 1

\(1^2-2^2+3^2-4^2+...-2004^2+2005^2\)

\(=1^2-2^2+3^2-4^2+...+2003^2-2004^2+2005^2\)

\(=-\left(2^2-1^2\right)-\left(4^2-3^2\right)-...-\left(2004^2-2003^2\right)+2005^2\)

\(=-\left[\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2004^2-2003^2\right)\right]+2005^2\)

\(=-\left[\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2004-2003\right)\left(2004+2003\right)\right]+2005^2\)

\(=-\left[1+2+3+4+...+2003+2004\right]+2005^2\)

\(=-\dfrac{2004.2005}{2}+2005^2\)

\(=2011015\)

:D