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Mathematics 10 Online
OpenStudy (anonymous):

What is 1-2+3........+2005-2006+2007

OpenStudy (anonymous):

1004, I think..

OpenStudy (anonymous):

I don't think that's it. It's either 2008 or 0, I think.

OpenStudy (anonymous):

it's 1004 :DD

OpenStudy (anonymous):

But how would that work? If you add the first and last term, second term, second to last term, etc. etc., you would get 2008, -2008, etc. etc.

OpenStudy (anonymous):

( sum of the odds) - (sum of the evens)

OpenStudy (anonymous):

That would be (some multiple of 2008)-(some multiple of -2008), though, right?

OpenStudy (anonymous):

what are saying? what multiple of 2008?

OpenStudy (anonymous):

Because like I said earlier, when you add the first term and last term, you get 2008. When you add the second and second to last term, you get -2008. When you add the third term and third to last term, you get 2008. When you add the fourth term to the fourth to last term, you get -2008. There is a pattern.

OpenStudy (anonymous):

@satellite73 Do you have any insight into this?

OpenStudy (anonymous):

If you try it with just 1-2+3.........-12+13, you get 7. That suggests the answer should be [greatest number in the problem] + 1, then that divided by 2. So @sourwing you are right.

OpenStudy (anonymous):

(1 + 2007) + (-2 + -2006) + (3 + 2005) + (-4 + -2004) + ... + (1003 + 1005) + (-1004)

OpenStudy (anonymous):

the sums add up to 2008 and so 2008 - 1004 = 1004

OpenStudy (anonymous):

the patter for the sum is like this: 2008 , -2008, 2008, -2008 so it's 2008 if n = odd, and -2008 if n = even look at the pair (1003 + 1005), n = odd, thus the sum is 2008 so 2008 - 1004 = 1004

OpenStudy (anonymous):

Ah, you're right, cause 1004 is in the middle...

OpenStudy (anonymous):

Thanks for your time and any troubles

OpenStudy (anonymous):

we no, -1004 is in the middle. The sum adds up to 2008. 2008 - 1004 = 1004

OpenStudy (anonymous):

yw

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