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You can put this solution on YOUR website!
2+4+6+ ... +2012 has 2012/2=1006 terms
\n" ); document.write( "So does -1+3+5+ ... +2011 and
\n" ); document.write( "so does 1+3+5+ ... +2011.
\n" ); document.write( "If you really meant
\n" ); document.write( "2+4+6+ ... +2012-(1+3+5+ ... +2011) it would be a really cute problem because
\n" ); document.write( "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
\n" ); document.write( "On the other hand,
\n" ); document.write( "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
\n" ); document.write( "is no fun.
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( "It's the same way when you add any sequence of evenly spaced numbers (called an arithmetic sequence).
\n" ); document.write( "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
\n" ); document.write( "(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= ...
\n" ); document.write( "So twice the sum would be 1006 times (2+2012). The sum is half of that, or 1006(2+2012)/2 \n" ); document.write( "