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You can put this solution on YOUR website!
It is true that we can take a calculator and add them up to get 63.
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\n" ); document.write( "It would be more interesting if we can sum the series:
\n" ); document.write( "(1+2-3)+(4+5-6)+....(3n-2 + 3n-1 -3n)
\n" ); document.write( "where there are n groups of triplets.
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\n" ); document.write( "We can use Gauss's method by writing the same series written backwards below the original series. The next step is to add up each group of 3 terms.
\n" ); document.write( "(1+2-3) +(4+5-6)+.... +(3n-2 +3n-1 -3n)
\n" ); document.write( "(-3n + 3n-1 +3n-2) +(-3n+3 +3n-4 +3n-5).... (-3 +2 +1 )
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\n" ); document.write( "3n-3 + 3n-3 +.... +3n-3
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\n" ); document.write( "(note: unfortunately proportionate spacing and removal of blanks does not let us match up each group, but you know what I mean!)
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\n" ); document.write( "Since there are n such groups, the sum is n(3n-3) for two series, therefore each series has a sum of 3n(n-1)/2.
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\n" ); document.write( "Check, for n=7 (given problem), we have sum = 3(7)(6)/2=63 as before.
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\n" ); document.write( "If you are curious as to what Gauss's method is, see:
\n" ); document.write( "http://nrich.maths.org/2478 \n" ); document.write( "