matrices -proof by induction

I was doing this at college today! you say 'let it be true for x = k, and then write it the same but with k instead of x. Then you say, show to be true for x = k +1; which when you write it, is the matrice you wrote with k (matrice ^k = [1 k 0 1]) *times* the original matrice Because: something to the power of ^k+1 is the same as something to the power of k *times* soemthing to the power of one (just the original number/matrice) (rules of indices) - then you just multiply them and should be able to factor k+1 where x was at the start, hence showing it's true for k and K+1 therefore all values of x

hope at least some part of that made any sense...!

hold up - you multiply [1, 1, 0, 1]^k by [ 1, 1, 0, 1]

which means you multiply [ 1, k, 0, 1] (because you said at the start let it be true that [1, 1, 0, 1]^k = [1, k, 0,1]) by [1, 1, 0, 1]
which should work out as [1, k+1, 0, 1]?

Yes you are right by showing it for $n=4$.

You need to show that $3^{k+1} > (k+1)^3$, given the assumption that $3^{k} > k^3$. What you've written doesn't show that.