SOLUTION: Hi Find the value of 1-2+3-4+5-6+.....+2007-2008+2009-2010+2011.

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Question 1209331: Hi
Find the value of 1-2+3-4+5-6+.....+2007-2008+2009-2010+2011.

Found 3 solutions by amarjeeth123, ikleyn, math_tutor2020:
Answer by amarjeeth123(570) About Me  (Show Source):
You can put this solution on YOUR website!
The given series is 1-2+3-4+5-6+.....+2007-2008+2009-2010+2011
The given series is equivalent to (1+3+5+........+2011)-(2+4+6+........+2010)
We can use Google Sheets and Excel to solve the given problem.
I have attached a snippet of the Excel sheet.
Answer=1006

Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.

This problem admits simple mathematical mental solution.

Group the terms in pairs

1-2+3-4+5-6+.....+2007-2008+2009-2010+2011 = (1-2)+(3-4)+(5-6)+.....+(2007-2008)+(2009-2010)+2011.


You have 2010/2 = 1005 pairs and the last unpaired term 2011.

Each pair produces -1, so the total expression is -1005+2011, and you can calculate it mentally.


ANSWER.   1006.


Solved  (mentally)


This problem is specially designed to be solved in this way.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 1006

Explanation

Tutor ikleyn has shown an efficient pathway. Perhaps the most efficient route.
I'll show an alternative method.

Let's place the terms of 1-2+3-4+5-6+...+2007-2008+2009-2010+2011 into a table

1-23-45-6...-20062007-20082009-20102011


Then we'll make a copy of this row and reverse it to place under the current terms
1-23-45-6...-20062007-20082009-20102011
2011-20102009-20082007-2006...-65-43-21


Add straight down to see each column adds to either 2012 or -2012.
More specifically the odd values add to 2012 while the even values add to -2012.

In the set {1,2,3,...,2010,2011} there are 2010/2 = 1005 even numbers.
Those even numbers are 2,4,6,...,2010.
This means that we have 1005 instances of the sum -2012 show up.
The other 2011-1005 = 1006 sums are 2012

We then have
1006*2012 + 1005*(-2012)
= 2012*(1006-1005)
= 2012*(1)
= 2012

This is not the final answer.
It would be nice if 1-2+3-4+...+2009-2010+2011 did evaluate to 2012.

However, when I made that 2nd table, where the bottom row is the reverse of the top row, I introduced a second copy of the sum. Thereby the result of adding everything in that 2nd table would be 2*S, where S = 1-2+3-4+...+2009-2010+2011

So we have to divide by 2 to correct this error.
2012/2 = 1006 is the final answer.


In other words, S = 1-2+3-4+...+2009-2010+2011 represents combining terms along the top row
S = 2011-2010+2009-...-4+3-2+1 also happens when we combine terms along the bottom row.
Add those equations straight down to arrive at 2S = 2012 which leads to S = 1006