Prove the following identity:
In any partition of the first 2N natural numbers into decreasing and increasing sequences of N members each, the sum of absolute values of the differences of the corresponding members of the two sequences is always N^2.
Source:
Savchev, Svetoslav; Andreescu, Titu. Mathematical Miniatures. Washington, D.C.
In any (N+, N) distribution of the 2N numbers, if we invert (or mirror) the sequence and consider the same distribution but inverted (N N+) we will have the same pairs of members for both distributions, (N+, N) and (N, N+), and the same sum value for both.
If we consider now both distributions together, as just one, what we really have is the whole increasing and decreasing sequences of the 2N numbers. So (2N+,2N).
The sum of the differences of the (2N+, 2N) distribution is a sum of N consecutive odd terms from 1 to (2N1), but repeated.
(f. ex. if N=7, (141)+(132)+(123)+(114)...= 13+11+9+7+5+3+1+1+3+5+7+9+11+13
This is wellknown sum whose result for N terms is the square N^2. As it is repeated the sum is 2N^2.
Then the puzzle's sum should be the half, so=N^2.
1 4 5 7 9
10 8 6 3 2
+
2 3 6 8 10
9 7 5 4 1
=
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Edited on March 5, 2016, 3:12 pm

Posted by armando
on 20160305 06:16:30 