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数学
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一列数—2009,—2005,—2001,—1997,..按照后一个数等于前一个数与4的和,则这列数的前几个个数和最小
人气:341 ℃ 时间:2020-09-04 04:19:59
解答
a1 = -2009,d=4
设前k项的和最小,则ak<0并且a(k+1)≥0
ak = a1 +(k-1)d = -2009+4k-4=-2013+4k<0
a(k+1) = a1 +kd = -2009+4k≥0
2009/4≤k<2013/4
502+1/4≤k<503+1/4
故前503个数的和最小
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