ttGWD1-28

summertao

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!

tangjinxiong

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

taotaobao

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

peiyangzhe

套公式 S=A1*N+（N-1）N*D/2

A1=100

N=101=（301-99）/2

D=2