First I would do the base case (the first value):Test n=1,111-6=5. 5 is divisible by 5 therefore true for n=1.
Now we assume true for n=k,11k-6 is divisible by 5.Next we test n=k+1,11k+1-6We can rearrange this into 1111k-6= 1011k+11k-6We know that for n=k the result is 11k-6 which we assume to be true so that part can be assumed to be true.The first part can be factorised into 5(2*11k) which is divisible by 5. Therefore we have shown that if true for n=k, true for n=k+1 and as we shown true for n=1 it must also be true for all natural numbers. So we have proved this through induction