Solution to Recurrence Relation I asked a question previously, about how to describe 
$$
f(n) = n^3
$$
As a recurrence relation. I was, quite rightly, given $a_1=1$ and $a_{n+1}=a_n+3n^2+3n+1$.
I have attempted to solve it, using forward substitution, but I'm having trouble.
I started out by assuming a solution to this recurrence relation was $n^3$. I then attempted a proof by induction that $a_n + 2 = (n+1)^3$
And now I am stuck out of my mind! Can anyone help me out here?

I don't understand your question. If you want to show that the solution to the recurrence $$a_{n+1} = a_n + 3n^2 + 3n + 1 \,\,\,\,\, a_1 = 1$$ is $a_n = n^3$, here is a way out.
We have
$$a_{n+1} + n^3 = a_n + \overbrace{n^3 + 3n^2 + 3n + 1}^{(n+1)^3} = a_n + (n+1)^3$$
Now if we call $b_n = a_n - n^3$, we have
$$b_{n+1} = b_n \,\, \text{ and } \,\, b_1 = a_1 - 1 = 0$$
Hence, we get $$b_{n+1} = b_n = b_{n-1} = \cdots = b_1 =0$$
This implies $$a_n = n^3$$
