2002^2-2001^+2000^2-1999^2+1998^2-…+2^2-1^2

2024-11-08 06:29:49
推荐回答(2个)
回答1:

两两组合
2002^2-2001^2+2000^2-1999^2+1998^2-…+2^2-1^2
=(2002+2001)(2002-2001)+(2000+1999)(2000-1999)+(1998+1997)(1998-1997)+......+(4+3)(4-3)+(2+1)(2-1)
=4003+3999+3995+......+7+3
=1001×(4003+3)/2
=2005003

注:项数=(4003-3)/4+1=1001,公差是4的等差数列

回答2:

2002^2-2001^2+2000^2-1999^2+1998^2-…+2^2-1^2
=(2002+2001)(2002-2001)+(2000+1999)(2000-1999)+(1998+1997)(1998-1997)+......+(4+3)(4-3)+(2+1)(2-1)
=(2002+2001)*1+(2000+1999)*1+(1998+1997)*1+...+(2+1)*1
=2002+2001+2000+1999+1998+1997+...+2+1
=(2002+1)*2002/2
=2003*1001
=2005003.

(说明:等差数列求和公式:(首数+尾数)*项数/2)