Question 938873
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 )
------------------------------------------------------------
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