2k³+15k²+37k+24/24= [(k+1)(2(k+1)²+9(k+1)+13)/24]
Expand the right-hand side: (2(k+1)^3+9(k+1)^2+13(k+1))/24.
The denominators are the same on each side so they can be cancelled out.
2((k+1)^3-k^3)+12(k+1)^2-12k^2-3(k+1)^2-3k^2+13k+13-37k-24=0;
2(3k^2+3k+1)+12(2k+1)-3(2k^2+2k+1)-24k-11=0;
6k^2+6k+2+24k+12-6k^2-6k-3-24k-11=0;
0=0; so the problem was in fact an identity, both sides being identically equivalent.
CHECK
Put k=0: 24=[2+9+13]. OK.
Put k=-1: -2+15-37+24=0. OK.