数学OG13题172不明白，能帮忙讲解一下吗

小皮球PAL · 发表于 2012-11-25 17:23:17

157. 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
答案Algebra Simplifying expressions; Arithmetic
Computation with integers
Th e given formula translates into 1 + 2 + … + n =
.
Th e sum of the even integers
between 99 and 301 is the sum of the even
integers from 100 through 300, or the sum of the
50th even integer through the 150th even integer.
To get this sum, fi nd the sum of the fi rst 150 even
integers and subtract the sum of the fi rst 49 even
integers. In symbols,
=
=
= 150(151) – 49(50)
= 50[3(151) – 49]
= 50(453 – 49)
= 50(404)
= 20,200

不明白为什么乘了个2，,谢谢讲解了

jsmart · 发表于 2012-11-25 17:49:33

99到301之间偶数之和=100+102+104+...+300=2(50+51+52+...+150)=2[(1+150)*150/2-(1+50)*50/2]

enyika2014 · 发表于 2014-10-18 01:01:34

gooooooooooooooooood

jessica_chen199 · 发表于 2014-10-30 08:37:53

99到301之间偶数之和=100+102+104+...+300=2(50+51+52+...+150)=2[(1+150)*150/2-(1+49)*49/2]
2是提取出来的，因为求得even integer，提取出来之后就有了连续整数，符合规律，套用公式。

数学OG13题172不明白，能帮忙讲解一下吗

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