(1)计算1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010(2)计算1+(-2)+3+(-4)+.+2009+(-2010)(3)总结(1)(2)两道题的规律
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 04:22:58
![(1)计算1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010(2)计算1+(-2)+3+(-4)+.+2009+(-2010)(3)总结(1)(2)两道题的规律](/uploads/image/z/16481-65-1.jpg?t=%281%29%E8%AE%A1%E7%AE%971%2B2%2B3%2B4%2B5-6%2B7%2B8%2B9%2B10...%2B2007%2B2008%2B2009%2B2010%282%29%E8%AE%A1%E7%AE%971%2B%EF%BC%88-2%EF%BC%89%2B3%2B%EF%BC%88-4%EF%BC%89%2B.%2B2009%2B%EF%BC%88-2010%EF%BC%89%EF%BC%883%EF%BC%89%E6%80%BB%E7%BB%93%EF%BC%881%EF%BC%89%EF%BC%882%EF%BC%89%E4%B8%A4%E9%81%93%E9%A2%98%E7%9A%84%E8%A7%84%E5%BE%8B)
(1)计算1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010(2)计算1+(-2)+3+(-4)+.+2009+(-2010)(3)总结(1)(2)两道题的规律
(1)计算1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010
(2)计算1+(-2)+3+(-4)+.+2009+(-2010)
(3)总结(1)(2)两道题的规律
(1)计算1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010(2)计算1+(-2)+3+(-4)+.+2009+(-2010)(3)总结(1)(2)两道题的规律
1、原式=(1+2010)+(2+2009)+.(一共有2010/2=1005对)
=2011*1005=2021055
2、原式=(1-2)+(3-4)+.(一共1005对).+(2009-2010)
=-1*1005=-1005
3、1、2式都是将许多的数字两两组合,使得每个组合的和相等来计算的.
(1)1+2+3+4+5-6+7+8+9+10...+2007+2008+2009+2010
=1/2*2010*(2010+1)=2021055
(2)1+(-2)+3+(-4)+......+2009+(-2010)
=-1-1-1-...-1
=-1*1005=-1005
(3) 1+2+3+...+n=1/2*n(n+1)
1-2+3-4+....-2n
=-1*n=-n
1 [(1+2010)×2010]/2
2 [(1+2009)×1005]/2 + {[(-2)+(-2010)]×1005}/2
利用的是等差数列求和公式 [(首项+末项)×项数]/2