Solution:
No pattern is evident in the first few terms, so we look for a formula for $a_{n}$. If we write $a_{n}=A n^{2}+B n+C+b_{n}$ and put $b_{n+2}=4 b_{n+1}-4 b_{n}$. Rewriting the original recurrence, we find
$$\begin{aligned} & A n^{2}+(4 A+B) n+(4 A+2 B+C)+b_{n+2} \\ & =4\left(A n^{2}+(2 A+B) n+(A+B+C)+b_{n+1}\right)-4\left(A n^{2}+B n+C+b_{n}\right)+n^{2} \\ & \quad=n^{2}+8 A n+(4 A+4 B)+4 b_{n+1}-4 b_{n} \end{aligned}$$
Solving, $A=1$, $B=4$, $C=8$. With this information, we can solve for $b_{0}=1$ and $b_{1}=6$. Since the characteristic equation of the recurrence of the $b_{i}$ is $x^{2}-4 x+4=(x-2)^{2}=0$, we have $b_{n}=(D n+E) \cdot 2^{n}$ for some constants $D$ and $E$. Using the known values $b_{0}$ and $b_{1}$, we compute $D=2$ and $E=1$, and finally
$$a_{n}=n^{2}+4 n+8+(2 n+1) \cdot 2^{n}$$
Now, taking modulo $100$, we have $a_{2006} \equiv 6^{2}+4 \cdot 6+8+13 \cdot 2^{2006} \pmod{100}$. Evidently $2^{2006} \equiv 0 \pmod{4}$, but by Euler's theorem $2^{\phi(25)} \equiv 2^{20} \equiv 1 \pmod{25}$, and so $2^{2006} \equiv 2^{6} \equiv 14 \pmod{25}$. Now the Chinese remainder theorem yields $2^{2006} \equiv 64 \pmod{100}$, and we compute $a_{2006} \equiv 36+24+8+13 \cdot 64 \equiv 0 \pmod{100}$.