一个序列由2010项组成.第一个项后的每个项比前一个项大1.2010项的总数是5307.从第一个项开始,最后第二个项加上每一个第二个项,总数是多少?
我们有一个由2010项组成的序列,其中第一个项后的每个项都比前一个项大1. 2010项的总数是5307. 我们需要从第一个项开始,最后用第二个项结尾,找到每个第二项的总和. 为了解决这个问题,我们标记了x_1,x_2,x_3,..., x_2009,x_2010. 让S是序列中的奇数项的总和,即:S = x_1 + x_3 + x_5 +  + x_2007 + x_2009 我们知道所有项的总数是5307,这意味着: x_1 + x_2 + x_3 +  + x_2009 + x_2010 5307. 接下来我们可以写出每一个奇数项的和下列奇数项的和. 简单的数组是100x=51,所以我们可以写出:S2 = 100x2 + 100x1 + 100x2 + 100x2 + 100x2 + 100x1 于每一个奇数项的和,所以我们可以写出:S2 = 100x2 + 100x1 + 100x2 + 100x2 + 100x1 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100x2 + 100).