1 | 2 | 3 | 4 | ... | n |

n | n-1 | n-2 | n-3 | ... | 1 |

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