limpid blaze
So one way of doing this q is using 6|k(k^2 + 5). However, I'm finding the proof of this method quite challenging. Can I do the same thing using k(k^2 + 5) = 6m, where m is an integer? If yes, can you show the steps on how to do the 2 steps after the base case.
quasi egretBOT
placid notch
So one way of doing this q is using 6|k(k^2 + 5). However, I'm finding the proof...
Induction is certainly the wrong way to go about proving this claim, but I digress.
For brevity, let P(n) be the statement that n(n^2+5) is a multiple of 6.
For induction, you have to show that P(1) is true.
Moreover, you have to show that if P(n) is true, then P(n+1) is true.
Star by writing out P(n+1), and then manipulate it so that you get n(n^2+5) and some additional terms. Those additional terms will be a multiple of 6, and n(n^2+5) will be a multiple of 6 by the induction hypothesis.
limpid blaze
.solved