SOLUTION: What is 1-2+3-4+5-6+...+2013? I have no idea how to do this problem. The answer is 1007 Any help is appreciated. Thank you!!

Question 1050865: What is 1-2+3-4+5-6+...+2013?
I have no idea how to do this problem. The answer is 1007
Any help is appreciated. Thank you!!
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
+...+....take a look at signs: positive, negative, positive,negative..
integers are positive and integers are negative
there is integers in all
since first and last integers, means there is integer than odd
since is not divisible by we can write it as where are odd integers and even integer
since even integers are negative

SUM OF EVEN NUMBERS:
SUM OF ODD NUMBERS:
since integers are negative, we subtract the sum of even from the sum of odd integers

........... and and we have
=
=
=
=
=
=

Answer by ikleyn(52777) (Show Source): You can put this solution on YOUR website!
.

Group the numbers in the series in pairs: 1 - 2 + 3 - 4 + 5 - 6 + . . . + 2013 = (1-2) + (3-4) + (5-6) + . . . + (2011-2012) + 2013 The last number, 2013, is without pair. Notice that every difference in parentheses is equal to -1. How many pairs do you have ? = 1006. So, you have the sum of 1006 terms of "-1" plus 2013, which is -1006 + 2013. Add the last two numbers. And compare with the answer.

Solved.

This problem is actually for low grade (young !) students who may not know about the sum of arithmetic progression.

This problem is to develop their combinatoric skills.

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