Question 538014: 2+4+6+...+2012-1+3+5+...+2011= ?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! 2+4+6+ ... +2012 has 2012/2=1006 terms
So does -1+3+5+ ... +2011 and
so does 1+3+5+ ... +2011.
If you really meant
2+4+6+ ... +2012-(1+3+5+ ... +2011) it would be a really cute problem because
2+4+6+ ... +2012-(1+3+5+ ... +2011)= 2+4+6+ ... +2012-1-3-5- ... -2011=(2-1)+(4-3)+(6-5)+ ... +(2012-2011)=1+1+1+ ...+1=1006
On the other hand,
2+4+6+ ... +2012-1+3+5+ ... +2011=(2+4+6+ ... +2012)-1+(3+5+ ... +2011)=1006(2+2012)/2-1+1005(3+2011)/2=1006*1007-1+1007*1005=1007(1006+1005)-1=1007*2011-1=2,2025,076
is no fun.
2+4+6+ ... +2012 is the sum of an arithmetic sequence and equals the number of terms (1006) times the average of the first and last terms.
3+5+ ... +2011 is the sum of an arithmetic sequence too and equals the number of terms (1005) times the average of the first and last terms.
It's the same way when you add any sequence of evenly spaced numbers (called an arithmetic sequence).
The reason why it is so is that if you added everything twice (to get twice the sum), you could pair the terms head to tail to get
(2+2012)+(4+2010)+(6+2008)+ ... +(2012+2), where you have 1006 sums that all add up to the same 2+2012=4+2010=6+2008= ...
So twice the sum would be 1006 times (2+2012). The sum is half of that, or 1006(2+2012)/2
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