Base Case: n=1, 1(1^2-1) = 0, divisible by 24. Inductive Step: Assume k(k^2-1) is divisible by 24 for odd k. Then (k+2)((k+2)^2-1) = (k+2)(k^2+4k+3) = k^3 + 6k^2 + 11k + 6 = (k^3-k) + 6k^2 + 12k + 6 = k(k^2-1) + 6(k^2 + 2k + 1) = k(k^2-1) + 6(k+1)^2. Since k is odd, k+1 is even, so (k+1)^2 is divisible by 4. Thus 6(k+1)^2 is divisible by 24. Therefore, (k+2)((k+2)^2-1) is divisible by 24.

Base Case: n=1, 1(1^2-1) = 0, divisible by 24. Inductive Step: Assume k(k^2-1) is divisible by 24 for odd k. Then (k+2)((k+2)^2-1) = (k+2)(k^2+4k+3) = k^3 + 6k^2 + 11k + 6 = (k^3-k) + 6k^2 + 12k + 6 = k(k^2-1) + 6(k^2 + 2k + 1) = k(k^2-1) + 6(k+1)^2. Since k is odd, k+1 is even, so (k+1) is divisible by 2. Thus 6(k+1)^2 is divisible by 24. Therefore, (k+2)((k+2)^2-1) is divisible by 24.

Base Case: n=1, 1(1^2-1) = 0, divisible by 24. Inductive Step: Assume k(k^2-1) is divisible by 24 for odd k. Then (k+2)((k+2)^2-1) = (k+2)(k^2+4k+3) = k^3 + 6k^2 + 11k + 6 = (k^3-k) + 6k^2 + 12k + 6 = k(k^2-1) + 6k^2 + 12k + 6. Since k is odd, k+1 is even, so (k+1) is divisible by 2. Thus 6(k+1)^2 is divisible by 24. Therefore, (k+2)((k+2)^2-1) is divisible by 24.

Base Case: n=1, 1(1^2-1) = 0, divisible by 24. Inductive Step: Assume k(k^2-1) is divisible by 24 for odd k. Then (k+2)((k+2)^2-1) = (k+2)(k^2+4k+3) = k^3 + 6k^2 + 11k + 6 = (k^3-k) + 6k^2 + 12k + 6 = k(k^2-1) + 6k^2 + 12k + 6. Since k is odd, k+1 is even, so (k+1)^2 is divisible by 4. Thus 6(k+1)^2 is divisible by 24. Therefore, (k+2)((k+2)^2-1) is divisible by 24.

The statement is not true, and thus cannot be proven via mathematical induction.