Three. (35 points) Let the sequence $\left\{U_{n}\right\}$ satisfy: $U_{\mathrm{b}}=1, U_{1}=3$. $U_{n+2}=4 U_{n+1}-U_{n}$. For non-negative integers $n$, determine the last two digits (i.e., the units and tens digits) of $2 U_{n}^{3}-U_{n}$.

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For $n=0,1,2 \cdots$, the first step is to divide $U_{n}$ by 4 to get a sequence with a period of 4: $1,3,3,1,1,3,3,1, \cdots$,
that is, when $n=4 k, 4 k+3$, it leaves a remainder of 1 when divided by 4;
when $n=4 k+1,4 k+2$, $U_{n}$ leaves a remainder of 3 when divided by 4.
The second step is to divide $U_{n}$ by 25 for $n=0.1,2, \cdots$, to get a sequence with a period of 15:
$$
1,3,11,16,3,21,6,3,6,21,3,16,11,3,1,1,3, \cdots
$$

From this, we can see: when $n=3 k+1$, $U_{n}$ leaves a remainder of 3 when divided by 25, and when $n \neq 3 k$ +1, $U_{n}$ leaves a remainder of 1 when divided by 5. $2 U_{n}^{2}-U_{n}-1=\left(U_{1}-1\right)\left(2 U_{n}^{2}+2 U_{n}+1\right)$, so when $U_{n}$ leaves a remainder of 1 when divided by 5, $U_{n}-1$ and $2 U_{n}^{2}+2 U_{n}+1$ are both divisible by 5, meaning $2 U_{3}-U_{n}$ leaves a remainder of 1 when divided by 25. When $U_{n}$ leaves a remainder of 3 when divided by 25, set $U_{n}=25 k+3$ and substitute into $2 U_{n}^{2}+2 U_{n}+1$ to find that $2 U_{n}^{2}+2 U_{n}+1$ is divisible by 25, so at this time $2 U_{n}^{3}-U_{n}$ is also divisible by 25 and leaves a remainder of 1. In summary, for any non-negative integer $n, 2 U_{n}^{3}-U_{n}$ always leaves a remainder of 1 when divided by 25. (This conclusion can also be verified by directly substituting all different remainders of $U_{n}$ when divided by 25, which are $1,3,6,11$, 21, into $2 U_{n}^{3}-U_{n}$)

The fourth step is to examine the remainder of $2 U_{n}^{3}$ when divided by 4. From the conclusion of the first step, it is directly derived that: when $n=4 k, 4 k+3$, $2 U_{n}^{3}-U_{n}$ leaves a remainder of 1 when divided by 4; when $n=4 k+1,4 k+2$, $2 U_{n}^{3}-U_{*}$ leaves a remainder of 3 when divided by 4.

From the conclusions of the third and fourth steps, we know: when $n=4 k, 4 k+3$, $2 U_{n}^{3}-U_{n}$ leaves a remainder of 1 when divided by 100, i.e., the last two digits are 01, when $n=4 k+1,4 k+2$, set $2 U_{x}^{3}-U_{n}=25 x+1=4 y$ +3 (where $x, y$ are positive integers), from this we solve $y=\frac{25 x-2}{4}=6 x$ $+\frac{x-2}{4}$, so $\frac{x-2}{4}=z$ is a non-negative integer, i.e., $x=4 z+2$. Substituting $2 U_{n}^{3}-U_{n}=25 x+1$ gives $2 U_{n}^{3}-U_{n}=25(4 z+2)+1$ $=100 z+51$. Therefore, we know that the last two digits of $2 U_{n}^{3}-U_{n}$ are 51.

In summary, when $n=4 k, 4 k+3$, the last two digits of $2 U_{n}^{3}-U_{n}$ are 01, and when $n=4 k+1,4 k+2$, the last two digits of $2 U_{n}^{3}-U_{n}$ are 51.