Question

Tính nhanh:

\(1^2-2^2+3^2-4^2+.....2002^2+2003^2-2004^2+2005^2\)

Answer 1

Ta có: 

1² - 2² + 3² - 4² + ... + 2003² - 2004² + 2005²

= 1² + (-2² + 3²) + (-4² + 5²) + ... + (-2002² + 2003²) +(-2004² + 2005²)

= 1 + (3² - 2²) + (5² - 4²) + ... + (2003² - 2002²) + (2005² - 2004²)

= 1 + (3 + 2)(3 - 2) + (5 + 4)(5 - 4) + .... + (2003 + 2002)(2003 - 2002) + (2005  + 2004)(2005 - 2004)

= 1 + 5.1 + 9.1 + .... + 4005 . 1 + 4009 . 1

= 1 + (5 + 9 + .... + 4005 + 4009)

= 1 + (4009 + 5)[(4009 - 5) : 4 + 1] : 2

= 1 + 4014 . 1002 : 2

= 1 + 2011014

= 2011015

Answer 2

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

\(=-\left(\left(2-1\right)\left(1+2\right)+...+\left(2005-2004\right)\left(2004+2005\right)\right)\)

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

\(=-\frac{2005\left(2005+1\right)}{2}=-2011015\)