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!!

Algebra ->  Permutations -> 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!!      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
1-2%2B3-4%2B5-6+...+2013....take a look at signs: positive, negative, positive,negative..
odd integers are positive and even integers are negative
there is 2013 integers in all
since first and last odd integers, means there is one+even integer less than odd
since 2013 is not divisible by 2 we can write it as 1007%2B1006+ where 1007 are odd integers and 1006 even integer
since even integers are negative


SUM OF EVEN NUMBERS: n%5Be%5D%28n%5Be%5D%2B1%29
SUM OF ODD NUMBERS: n%5Bo%5D%5E2
since even integers are negative, we subtract the sum of even from the sum of odd integers

n%5Bo%5D%5E2-n%5Be%5D%28n%5Be%5D%2B1%29...........n%5Be%5D=1006 and n%5Bo%5D=1007 and we have
=+1007%5E2-1006%281006%2B1%29
=1007%5E2-1006%281006%2B1%29
=1014049-1006%5E2-1006
=1014049++-1012036-1006
=1014049++-1013042
=1007

Answer by ikleyn(52777) About Me  (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 ?  2012%2F2 = 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.