Question

求和：1²-2²+3²-4²+5²-6²+…+99²-100²

如题 特殊数列求和

Answer 1

1²-2²=(1-2)×(1+2)=-(1+2)
3²-4²=(3-4)×(3+4)=-(3+4)
以此类推,
原式=-(1+2+3+4+...+100)=-5050

Answer 2

原式=(1²-2²)+(3²-4²)+……+(99²-100²)
=(1+2)(1-2)+(3+4)(3-4)+……+(99+100)(99-100)
=(1+2)*(-1)+(3+4)*(-1)+……+(99+100)*(-1)
第二个括号都是-1
所以=(-1)*(1+2+3+4+……+99+100)
=-5050

Answer 3

1²-2²+3²-4²+5²-6²+…+99²-100²
=(1+2)(1-2)+(3+4)(3-4)+...+(99+100)(99-100)
=-1-2-3-4-...-99-100
=-(1+2+...+100)
=-100×(1+100)÷2
=-5050;

Answer 4

1²-2²+3²-4²+5²-6²+…+99²-100²
=(1²-2²)+(3²-4²)+(5²-6²)+…+(99²-100²)
=(1-2)(1+2)+(3-4)(3+4)+…+(99-100)(99+100)
=-1-2-3-4-……-99-100
=-5050