5<SUP>2n</SUP> - 1 = 24p for n > 0, where p is some integer
i) test for n = 1
LHS = 25 - 1
= 24 which is dividisble by 24 therefore true for n = 1
ii) assume true for n = k
i.e. 5<SUP>2k</SUP> - 1 = 24q where q is some integer
iii) prove for n = k+1
5<SUP>2(k+1)</SUP> - 1 = 24q
LHS = 5<SUP>2(k+1)</SUP> - 1
= 5<SUP>2k+2</SUP> - 1
=5<SUP>2</SUP>x5<SUP>2k</SUP> - 1 ----- (1)
from assumption; 5<SUP>2k</SUP> - 1 = 24q therefore
5<SUP>2k</SUP>= 24q +1 ------- (2)
substitute (2) into (1)
LHS = 5<SUP>2</SUP>(24q +1) - 1
= 25 x 24q + 25 - 1
= 25 x 24q + 24
= 24 (25q + 1)
= 24q since (25q + 1) is some integer
therefore true for n = k+1
iv) since result is true for n = 1, and then from step 3 is true for n = 1+1 = 2, and then for n = 2+1 = 3 and so on for all positive integral of n.
yea step (iv) has never made sense to me but my teacher told me write that hahah. but everything else should be right but check over it just to be sure.
hope that helps