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.