Let's denote the largest odd divisor of a positive integer n by lod(n).
Thus lod(1)=1; lod(72)=9; lod(2k+1)=2k+1; lod(2^k)=1.
Prove the following statement:
Sum of all values of lod(k), (k>1) from k=n+1 to k=2n, inclusive, equals n2.
Example: take n=7: lod(8,9,10,11,12,13,14)= (1,9,5,11,3,13,7), sum of the values within the last pair of brackets is 49, indeed.
Source: Shown to me as a trick i.e. "the wizard" guesses the result a priori.
(In reply to One type of proof
Proving by induction, wherever applicable is a legal and reliable method. Nobody needs another proof.
I've erroneously introduced an erroneous and redundant constraint i.e. k>1, not being aware that the formula applies to k=1 as well.
Sorry about that.
I second SH's motion: your presentation is short and elegant.