Question

Tính \(K=1^2-2^2+3^2-4^2+5^2-6^2+...-2004^2+2005^2\)

Answer 1

Ta có: \(K=1^2-2^2+3^2-4^2+......+2005^2\)

\(\Rightarrow K=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+.....\) \(+\left(2005^2-2004^2\right)\)

\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)\)\(+......+\left(2005-2004\right)\left(2005+2004\right)\)

\(\Rightarrow K=1+5+9+13+.....+4009\)

Số số hạng trong tổng K là \(\frac{4009-1}{4}+1=1003\)

\(\Rightarrow K=\frac{\left(4009+1\right).1003}{2}=2005.1003\) = 2011015