Mohab wrote: > not so, what if you go negative? > (k-1)^2 + (k-1)
Factorise k^2 + k as k(k+1). Substituting k-1 for k gives (k-1)k, and the previous proof that since one of these is odd and the other even therefore the product is even holds. Of course this form of proof could be used to prove the theorem for all k, so it really shows up how proof by induction is unneccessary for this problem, but it would still be a valid proof by induction would it not?