求(1^2+3^2+5^2+…+99^2)-(2^2+4^2+6^2+…+100^2)的值

原式=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+.......+(99^2-100^2)=(1+2)(1-2)+(3+4)(3-4)+(5+6)(5-6)+...+(99+100)(99-100)=-1-2-3-4-5-6-.....-99-100=-5050

(1^2+3^2+5^2+…+99^2)-(2^2+4^2+6^2+…+100^2)
=（1^2-2^2）+（3^2-4^2）+…+(99^2-100^2)
=-(1+2+…+100)
=-(1+100)*100/2=-5050

1^2-2^2=-3
3^2-4^2=-7
5^2-6^2=-11
....
-3 -7 -11 -15 - ... -199 = -5050

-(1+2 + 3+4 + 5+6 + .... +100) = -5050