用平方差公式
1-2^2+3^2-4^2+5^2-6^2+……+2007^2-2008^2
=(1+2)(1-2)+(3+4)(3-4)+……+(2007+2008)(2007-2008)
=(-1)*(1+2+3+4+……+2008)
=-[(1+2008)+(2+2007)+……+(1004+1005)]
=-2009*1004
=-2017036
1-2^2+3^2-4^2+5^2-6^2+……+2007^2-2008^2
=(1-2)(1+2)+(3-4)(3+4)+……+(2007-2008)(2007+2008)
=-3-7-……-4015
=-1004*(4015+3)/2
=-2017036
1-2^2+3^2-4^2+5^2-6^2+……+2007^2-2008^2
=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+.....(2007-2008)(2007+2008)
=-sigema [n+n+1)]
此处(n从1到2007)
结果自己求
可看成1^2-2^2+3^2-4^2+5^2-6^2+……+2007^2-2008^2
=(1+2)(1-2)+(3+4)(3-4)+(5+6)(5-6)+……+(2007+2008)(2007-2008)
=-(1+2+3+4+5+6+……+2007+2008)
=-2017036
两两结合:
(1+2)(1-2)+(3+4)(3-4)+...+(2007+2008)(2007-2008)
=-(1+2+3+...+2008)
=-2009*1004