Question

Tính

A = \(1^2-2^2+3^2-4^2+...+99^2-100^2\)

Answer 1

\(A=1^2-2^2+3^2-4^2+...+99^2-100^2\)

\(A=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(99^2-100^2\right)\)

\(A=\left(1+2\right)\left(1-2\right)+\left(3+4\right)\left(3-4\right)+...+\left(99+100\right)\left(99-100\right)\)

\(A=\left(1+2\right)\cdot\left(-1\right)+\left(3+4\right)\cdot\left(-1\right)+...+\left(99+100\right)\cdot\left(-1\right)\)

\(A=\left(-1\right)\cdot\left(1+2+3+...+100\right)\)

\(A=\left(-1\right)\cdot\dfrac{100\cdot\left(100+1\right)}{2}=\left(-1\right)\cdot5050=-5050\)

Answer 2

=(1²-100²)+(99²-2²)+...+(50²-51²)

=(1-100)(1+100)+(2-99)(2+99)+...+(50-51)(50+51)

=-99.101-97.101-...-1.101

=-101(1+3+5+...+99)

=-101\(\dfrac{\left(99+1\right).50}{2}\) = -252500