The sum of 1, 2, ..., n (digits from one to n) is found from writing them a second time, in reverse order, below themselves.


Adding the two lines gives n+1 in each of the n places. If those values are also added we get the sum twice, from the way we wrote the elements. A way to write this in symbols is first n(n+1) for the sum twice [n times (n+1)]. and then for the sum, half that.

Apply the above to the coin problem, means evaluate 10(10+1)/2. Hence:

Add all the coins together and weigh them. The result is 55 [ 10(11)/2 ] plus or minus error caused by the fake coins. Error is found from subtracting 55 from the weighing result. The unknown number of fake coins is error divided by the (known) size (magnitude) of difference between good and fake. Whether fakes are heavy or light also is found by the sign of error. (Heavy, plus; light, minus.)