The first thing to do is try a few values. We note that each u(n) is determined by the values u(0),...,u(n-1) so there is a unique solution.

We find that the first few values are 3, 16, 80, 400, 2000, 10000, ... and that for n > 1 they appear to satisfy u(n) = 16.5^(n-2).

Now let's put that back into the original. Is it true that 16.5^(n-2) is equal to the sum of (4k-1).16.5^(n-k-2) for k =1..(n-2) plus (4(n-1)-1).3 ?