Đặt \(A=-1^2+2^2-3^2+4^2-\cdots+\left(-1\right)^{n}\cdot n^2\)
TH1: n=2k
=>\(A=-1^2+2^2-3^2+4^2+\cdots+\left(-1\right)^{2k}\cdot n^2\)
\(=-1^2+2^2-3^2+4^2+\cdots+n^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+\cdots+\left(n-n+1\right)\left(n+n-1\right)\)
=1+2+3+4+...+n-1+n
\(=\frac{n\left(n+1\right)}{2}\)
TH2: n=2k-1
=>\(A=-1^2+2^2-3^2+4^2+\cdots+\left(-1\right)^{2k-1}\cdot n^2\)
\(=-1^2+2^2-3^2+4^2+\cdots-n^2\)
=(2-1)(2+1)+(4-3)(4+3)+...+(n-1-n+2)(n-1+n-2)-n^2
=1+2+3+4+...+(n-2)+(n-1)-n^2
\(=\frac{\left(n-1+1\right)\left(n-1\right)}{2}-n^2=\frac{n\left(n-1\right)}{2}-n^2=\frac{n\left(n-1\right)-2n^2}{2}=\frac{-n^2-n}{2}\)
