What would be a closed form relating this recurrence relation $W_n$ to $W_1$?
If $$W_n = \frac{z_n - 4W_{n-1}}{4n}$$
I keep nested fractions, and I'm not sure how to simplify without algebra being all over the page.
This is basically me forming a relationship with my current semester's WAM (average mark per subject) with the previous semester's, so $z_n$ here is just basically the accumulative marks of the current semester over 4 subjects, so it might have a sequence of $z_k$ in the final result.
I hypothesise that
$$W_1 = \frac{1}{4} \left(\sum_{k=2}^{n-1} \: (-1)^k (n-1)! \: z_k \right) + n!\:(-1)^{n+1} \: W_{n}$$ Can someone confirm this?
By multiplying $$W_n = \frac{z_n - 4W_{n-1}}{4n}$$ by $(-1)^nn!$, one gets $$ (-1)^nn!\cdot W_n = (-1)^n\frac14(n-1)!\cdot z_n + (-1)^{n-1}(n-1)! \cdot W_{n-1} $$ or$$ (-1)^nn!\cdot W_n -(-1)^{n-1}(n-1)! \cdot W_{n-1} = (-1)^n\frac14(n-1)!\cdot z_n $$ then summing from $n=1$ to $n=N$, on the left hand side terms telescope, one has $$ (-1)^N N!\cdot W_N -W_{0}=\sum_{n=1}^N(-1)^n\frac14(n-1)!\cdot z_n $$ giving