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.
Start with a slightly different problem.
The sum of the absolute differences from n of the numbers from 1 to 2n is obviously n^2; the sum of two of each kind, plus n = n(n1)+n = n^2. Call these the absdifs.
Say n=5, and say that we abandon the constraint as to decreasing and increasing sequences of n members to get, say:{8,5,9,4,1}, {7,6,10,3,1} with an absolute value sum of 5. What happened to the absdifs totalling 25?
The answer is that they are still there, but are being cancelled out, because the pairwise absdif signs are the same: {3,0,4,1,4}, {2,1,5,2,3}: absdifs 25, absolute value sum 5.
However, if we insist on decreasing and increasing sequences, then we are guaranteed that the signs of the absdifs will be different (except for the one zero). And as long as we ensure this, the sequences don't even need to be decreasing and increasing sequences: e.g. {7,1,9,5,4},{2,6,3,10,8} with absdifs of {2,4,4,0,1}{3,1,2,5,3}
Edited on March 5, 2016, 6:23 am

Posted by broll
on 20160305 01:22:05 