Question 938873: what is the sum of the series
1+2-3+4+5-6+7+8-9...-18+19+20-21
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! It is true that we can take a calculator and add them up to get 63.
It would be more interesting if we can sum the series:
(1+2-3)+(4+5-6)+....(3n-2 + 3n-1 -3n)
where there are n groups of triplets.
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.
(1+2-3) +(4+5-6)+.... +(3n-2 +3n-1 -3n)
(-3n + 3n-1 +3n-2) +(-3n+3 +3n-4 +3n-5).... (-3 +2 +1 )
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3n-3 + 3n-3 +.... +3n-3
(note: unfortunately proportionate spacing and removal of blanks does not let us match up each group, but you know what I mean!)
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.
Check, for n=7 (given problem), we have sum = 3(7)(6)/2=63 as before.
If you are curious as to what Gauss's method is, see:
http://nrich.maths.org/2478
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