[求助]PREP里的一道数学，题目都看不懂哦～～

lulu0829

For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variation in sign for the sequence 1, -3, 2, 5, -4, -6.

A one, B two, C three, D four, E five.

KEY:C

哪位NN 来帮忙解答一下～～我题目都没看懂什么意思哦 ～～～THX~~~

leo-liu

1*-3<0

-3*2<0

2*5>0

5*-4<0

-4*-6>0

有三个小于0的，所以选C

lulu0829

3Q~~~~

但是看不懂这个 the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative

能不能帮我翻译下？THX~

iamliz

"the number of variations in sign" 这东西的定义是 "在这个序列中, 有多少对相邻的两个数的乘积<0"

题目是问, 在序列 1, -3, 2, 5, -4, -6 中, "the number of variations in sign"是多少, 即"有多少对相邻的两个数的乘积<0"?

{1,-3}{-3,2}{5,-4}

[此贴子已经被作者于2008-10-11 17:29:59编辑过]