求OG12 上的一道数学题

sissi8466 · 发表于 2011-3-16 09:10:41

OG 12--157: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

我看了OG 的解释，有点不明白，为什么在算偶数和用公式时前面要乘以2？
请各位大侠帮忙讲解下

sdcar2010 · 发表于 2011-3-16 09:21:44

The sum of all even integers between 99 and 301: 100, 102, 104, ..., 296, 298, 300

Sum = (100+300)*100/2 = 20,000

There are in total 100 even numbers and they pair up for 50 pairs.

sissi8466 · 发表于 2011-3-16 09:38:24

你的算式我有点不明白：是用的n(n+1)/2 的公式吗？100＋300是什么意思啊

不过OG 的答案是20200

sdcar2010 · 发表于 2011-3-16 09:51:53

I missed the median - 200

101 integers between 100 and 300. Two hundred is the "odd man" in the pairing parties.

sdcar2010 · 发表于 2011-3-16 09:55:53

If you really want to use the equation n(n+1)/2, you can do it this way:

100, 102, 104, ..., 296, 298, 300
if all the above numbers are divided by 2, you get:
50, 51, 52 . . . 148, 149, 150

The sum from 1 to 150 is 150*(150 + 1)/2.

The sum from 1 to 49 is 49*(49+1)/2

The final answer for the question is 2*[ 150*(150 + 1)/2 - 49*(49+1)/2 ]

sissi8466 · 发表于 2011-3-16 22:29:40

thank you!

求OG12 上的一道数学题

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