Let 52n+2−24n−25 be denoted by f(n);
then f(n+1)=52n+4−24(n+1)−25
=52⋅52n+2−24n−49;
∴f(n+1)−25f(n)=25(24n−25)−24n−49
=576(n+1).
Therefore if f(n) is divisible by 576, so also is f(n+1); but by trial we see that the theorem is true when n=1, therefore it is true when n=2, therefore it is true when n=3, and so on; thus it is true universally.
The above result may also be proved as follows:
52n+2−24n−25=25n+2−24n−25
=25(1+24)n−24n−25
=25+25.n.24+M(242)−24n−25
=576n+M(576)
=M(576).