GWD-1-28

Ferdinand

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

Answer: B

蜂鸣

这个公式[n(n+1)]/2其实就是等差数列求和公式。也就是首项1加上末项n乘以个数n除以2.
99到301一共有101个偶数，具体如下：
100-199 50个
200-299 50个
300-301 1个
共计101个。
首项是100，末项是300，个数是101个，所以是(100+300)101/2=20200 所以选B