Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example:
$2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$
Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogenous equation is of the form $f^h(n) = c_1+c_2n$. But how does one deal with the constant 3 in this case?

Let $f_p = A + Bn + Cn^2$.
$$
\begin{cases}
2f(n) = 2A +2Bn + 2Cn^2\\
-f(n+1) = -A - B(n+1) - C(n+1)^2\\
-f(n-1) = -A - B(n-1) - C(n-1)^2
\end{cases}
\quad \Rightarrow
$$
$$
3 = Cn^2 - C(2n^2 + 2) = -2C \quad \Rightarrow \quad C = -\frac{3}{2}
$$
Thus, 
$$
f(n) = C_1 + C_2n -\dfrac{3n^2}{2}
$$
Now use the given initial conditions to find the constants $C_1$ and $C_2$.
