Sorry, but I couldn't understand your questions in the other thread...
so for 1)
the proposition is: for
so you try for (n = 2):
which is true for k = 1.
Then you accept the hypothesis: for for all n. And try and show that:
sub assumption, ie.
this means that you're only interested in showing that:
and then for you see that if n is even that 3 times an even number is divisble by 6 (3n). If n is odd you see that the (n + 1) term is even and so once again 3 times an even number (greater than 2 (lol)) is also divisible by 6. (You should probably write this up in a more formal way.) Finished.