Question

[zZz Phan Cả Phát zZz]

Tính nhanh 

\(A=\left(2004^2+2002^2+2000^2+...+4^2+2^2\right)-\left(2003^2+2001^2+1999^2+...+5^2+3^2+1\right)=\)

Answer 1

[kaitovskudo]

A=(2004²-2003²)+(2002²-2001²)+...+(2²-1²)

A=(2004-2003)(2004+2003)+(2002-2001)(2002+2001)+...+(2-1)(2+1)

A=2004+2003+2002+2001+...+2+1

A=(2004+1).2014:2

A=2029105