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标题: ttGWD1-28 [打印本页]

作者: xiaobenzhu 时间: 2006-9-26 12:11
标题: ttGWD1-28

Q28:

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?

A. 10,100

B. 20,200

C. 22,650

D. 40,200

E. 45,150

I cannot get the right answer B. Please help!

作者: hhhhhhzzzzzz 时间: 2006-9-26 12:48
first step: (99+301)*x/2
second step, you want to know what x is, which means how many numbers there, 301 is the 151st even intergral, 99 is the 50th, then x=151-50+1=102
third step plug in x=102 back to the equation you will get the answer B

作者: sunnyever 时间: 2007-3-15 15:26
这题的前半句话是没用么？好像用不上啊

作者: zifengling27 时间: 2007-3-19 02:32

我觉得应该是这么算的吧

题目是问从99到301的所有偶数之和,算的时候应该把100与300相加,102与298相加,104与296相加……这样依次相加总共会得出50个400，再加上200没有相对应的数相加，所有总和等于：50X400+200=20200

作者: catherine_fo 时间: 2007-4-2 16:19

套公式 S=A1*N+（N-1）N*D/2

A1=100

N=101=（301-99）/2

D=2

作者: joy110211 时间: 2007-4-16 08:11

套公式 S=A1*N+(N-1)N*D/2

A1=100

N=101=（301-99）/2

D=2

請問可以解釋一下公式嗎? 不太明白.

照你的公式, 只需要A1*N + (N-1)N 就可以得出答案20,200,

但你後面(*D/2)是什麼意思呢?

作者: windswings 时间: 2007-4-16 11:39
（（（301-99）/2）*（301+99））/2=20200其中两个括号不影响计算结果和顺序，只是体现下想法

作者: app228 时间: 2007-7-4 10:30

i think it should be

(200*(200+2))/2 = 20200

the 200 is the number of integers between 100 to 300

作者: 荷叶田田 时间: 2008-1-24 16:55
我也是用2楼的方法算的，结果得到的结果是20400....

作者: sandywood 时间: 2008-6-8 04:29

看不懂以上各位的解法，能不能说的详细点？

特别是那n(n+1)/2在解题时有什么用啊?

作者: chaosplus 时间: 2009-7-27 22:26
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]

作者: paul2006 时间: 2009-8-4 18:52

作者: 双鱼游 时间: 2010-9-14 18:25
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.

作者: 疯婆子颠三 时间: 2013-7-4 10:56
题目前半部分是没用的嚒。。。这种有干扰性的题目多么？？

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