Question

tính tổng S=2006²-2005²+2004²-2003²+...+4²-3²+2²-1²

Answer 1

\(S=2006^2-2005^2+2004^2-2003^2+....+2^2-1^2\)

\(=\left(2006-2005\right)\left(2006+2005\right)+\left(2004-2003\right)\left(2004+2003\right)+...\left(2-1\right)\left(2+1\right)\)

\(=2006+2005+2004+....+2+1\)

\(=\frac{2006\left(2006+1\right)}{2}=2013021\)

Answer 2

sao có 2006+2005+2004+2003+...+2+1 zay