ttGWD1-28

chaosplus

说白了就是求公差为2的等差数列，a1=100,an=300,n=101
n=((301-99)+1)/2=101余1，说明101个奇偶对，余下一个基数301，所以n=101
(a1+an)*101/2=20200

PS：题目中前半句话如何利用没想明白～

[此贴子已经被作者于2009/7/27 22:44:26编辑过]

paul2006

??

双鱼游

The even integers between 99 and 301 are 100, 102, 104………..300.
Their sum is 100+102+104…….+300 = 2*(50+51+……..+150).

First calculate the sum of all integers between 1 and 150 inclusive which is 150*(151)/2 = 11325.

Then calculate the sum of integers from 1 to 49 inclusive which is 49*50/2 = 1225.

The difference of the two sums which is 11325 - 1225 =10100 gives the sum of integers between 50 and 150 inclusive.

Now 2*10100 = 20200 gives the value of 2*(50+51+……..+150).

The correct answer is hence b.

疯婆子颠三

题目前半部分是没用的嚒。。。这种有干扰性的题目多么？？