Solve recurrence equations-homework extras Extras from my homework. The first one should be easier, but still hard enough.
1) $a_{n+3}-(3/2)a_{n+2}-a_{n+1}-(1/4)a_n=0$
2) $a_{n+3}-3a_{n+2}-3a_{n+1}+a_n=n^2+2^n$

First problem
Hint:
For the first one, you can use the same trick of the fibonacci numbers.
The characteristic polynomial for this question is:
$$q^3-\frac{3q^2}{2}-q-\frac{1}{4}=0\;.$$
The second Mathematica says:
Solution:
$$\begin{align*}
a(n)&= c_1 \left(2-\sqrt{3}\right)^n+c_2
   \left(2+\sqrt{3}\right)^n+c_3 (-1)^n\\\\
&+\frac{-24 \sqrt{3} \zeta
   \left(-2,\frac{n}{2}\right)+48 \zeta \left(-2,\frac{n}{2}\right)+24
   \sqrt{3} \zeta \left(-2,\frac{n+1}{2}\right)-48 \zeta
   \left(-2,\frac{n+1}{2}\right)}{9 \left(\sqrt{3}-1\right)^3
   \left(1+\sqrt{3}\right)}\\\\
&+\frac{12 \sqrt{3} n^2-24 n^2-30 \sqrt{3} n+60
   n-2^{n+3}+\sqrt{3} 2^{n+2}+27 \sqrt{3}-54}{9 \left(\sqrt{3}-1\right)^3
   \left(1+\sqrt{3}\right)}
\end{align*}$$
