Question 128, Define the sequence $\left\{a_{n}\right\}: a_{0}=a, a_{n+1}=2 a_{n}-n^{2}(n \geq 0)$, try to find all real numbers $a$, such that for any non-negative integer $\mathrm{n}$, we have $\mathrm{a}_{\mathrm{n}} \geq 0$.
Given the condition: $\frac{a_{n+1}}{2^{n+1}}=\frac{a_{n}}{2^{n}}-\frac{n^{2}}{2^{n+1}}$, hence
$$
\begin{array}{l}
\frac{a_{0}}{2^{0}}-\frac{a_{n}}{2^{n}}=\sum_{k=0}^{n-1}\left(\frac{a_{k}}{2^{k}}-\frac{a_{k+1}}{2^{k+1}}\right)=\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k+1}} \\
\Rightarrow \frac{a_{n}}{2^{n}}=a-\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k+1}}
\end{array}
$$

Let $S_{n}=\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k+1}}$, then $2 S_{n}=\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k}}$, thus we have:
$$
\begin{array}{l}
S_{n}=2 S_{n}-S_{n}=\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k}}-\sum_{k=0}^{n-1} \frac{k^{2}}{2^{k+1}}=-\frac{(n-1)^{2}}{2^{n}}+\sum_{k=0}^{n-2} \frac{(k+1)^{2}-k^{2}}{2^{k+1}} \\
=-\frac{(n-1)^{2}}{2^{n}}+\sum_{k=0}^{n-2} \frac{2 k+1}{2^{k+1}}=-\frac{(n-1)^{2}}{2^{n}}+\sum_{k=0}^{n-2} \frac{k}{2^{k}}+\sum_{k=0}^{n-2} \frac{1}{2^{k+1}} \\
=-\frac{(n-1)^{2}}{2^{n}}+\sum_{k=0}^{n-2} \frac{k}{2^{k}}+1-\frac{1}{2^{n-1}} \\
=1-\frac{(n-1)^{2}+2}{2^{n}}+\sum_{k=0}^{n-2} \frac{k}{2^{k}}
\end{array}
$$

Let $T_{n}=\sum_{k=0}^{n-2} \frac{k}{2^{k}}$, then $2 T_{n}=\sum_{k=0}^{n-2} \frac{k}{2^{k-1}}$, thus we have:
$$
\begin{array}{l}
T_{n}=2 T_{n}-T_{n}=\sum_{k=0}^{n-2} \frac{k}{2^{k-1}}-\sum_{k=0}^{n-2} \frac{k}{2^{k}}=-\frac{n-2}{2^{n-2}}+\sum_{k=0}^{n-3} \frac{(k+1)-k}{2^{k}} \\
=-\frac{n-2}{2^{n-2}}+\sum_{k=0}^{n-3} \frac{1}{2^{k}}=-\frac{n-2}{2^{n-2}}+2-\frac{1}{2^{n-3}}=2-\frac{n}{2^{n-2}}
\end{array}
$$

Substituting (3) into (2) gives $S_{n}=3-\frac{n^{2}+2 n+3}{2^{n}}$, and substituting this into (1) gives:
$$
a_{n}=2^{n}\left(a+\frac{n^{2}+2 n+3}{2^{n}}-3\right)
$$

Clearly, the condition is satisfied for $\mathrm{a} \in[3,+\infty)$.