ttGWD1-28

zm_patrickS

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!

natalietong

i think it should be

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

the 200 is the number of integers between 100 to 300

huangluyao

（（（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

Kath1986

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?

xiaotangyu

套公式 S=A1*N+（N-1）N*D/2

A1=100

N=101=（301-99）/2

D=2

yoyo9012

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

tingting520

这题的前半句话是没用么？好像用不上啊

besttys

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