There are many ways to prove this; let's look at the additive
identity as it applies to rown and rown+1.
1 n ...
n 1
/ \/ \ / \/
\
1 n+1
... n+1 1
The lines indicate that the 1 in rown is
used in 2 different sums, the 1 and the n+1 in rown+1;
likewise every element in rown is "standing on the shoulders"
of two different elements in rown+1, and is included in
the summing process of both. That means the sum of rown+1
must be twice the sum of rown, since every element of rownis represented in exactly two elements of rown+1.
