Challenging math question (9th grade math)?

what is the diffrence between the sum of the first 2004 positve integers and the sum of the next 2004 postive?


A) 2004 B) 4008 C) 2004^2 D) 4008^2





This was my math homework and i am totally lost but i don't want to loose the points for the problem so can someone please help me. If you can solve this please try to explain how you sound this answer. Thanx

Challenging math question (9th grade math)?
I'd say C) 2004^2





Use analogical sum


1)sum of first 3 and the next 3


1 + 2 + 3 = 6


4 + 5 + 6 = 15


15 - 6 = 9 = 3^2





2)sum of first 4 and the next 4


1 + 2 + 3 + 4 = 10


5 + 6 + 7 + 8 = 26


26 - 10 = 16 = 4^2


.


.


.


.


until n


(all got same result, the difference is n^2)
Reply:The sum of n numbers is n(n+1)/2


so, the sum of the first 2004 integers is


2004(2005)/2 = 1002*2005


the sum of the next 2004 integers is the sum of the first 4008 integers - sum of the first 2004 integers


(4008(4009)/2) - 1002*2005 = 2004*4009 - 1002*2005


the difference between the 2 is





2004*4009 - 1002*2005 - 1002*2005


which is:


2004*4009 - 2004*2005, or


2004(4009 - 2005) = 2004(2004) = 2004^2





so, the answer is C

aster