The ninetieth problem: Given a positive integer $\mathrm{n}$, let $\mathrm{P}(\mathrm{x})$ and $Q(\mathrm{x})$ be real-coefficient polynomials with degrees not exceeding $\mathrm{n}$, satisfying $x^{n+1} P(x)+(x+1)^{n+1} Q(x)=1$. Try to find $Q(x)$, and determine the value of $Q\left(-\frac{1}{2}\right)$.
The 90th problem,
Solution method one: According to the condition, we know that $x^{n+1} \mid\left((x+1)^{n+1} Q(x)-1\right)$, so $x=0$ is a root of multiplicity $n+1$ of the polynomial $R(x)=(x+1)^{n+1} Q(x)-1$. Therefore, we know that $R(0)=0$, which means $Q(0)=1$; and $x=0$ is a root of multiplicity $n$ of the polynomial
$$
\begin{array}{l}
R^{\prime}(x)=(n+1)(x+1)^{n} Q(x)+(x+1)^{n+1} Q^{\prime}(x) \\
=(x+1)^{n}\left((n+1) Q(x)+(x+1) Q^{\prime}(x)\right)
\end{array}
$$

Thus, we know
$$
x^{n} \mid\left((n+1) Q(x)+(x+1) Q^{\prime}(x)\right)
$$

Let $Q(x)=\sum_{i=0}^{n} a_{i} x^{i}$, then $Q(0)=a_{0}=1$. And
$$
\begin{array}{l}
(n+1) Q(x)+(x+1) Q(x) \\
=\sum_{i=0}^{n}(n+1) a_{i} x^{i}+(x+1)\left(\sum_{i=1}^{n} i a_{i} x^{i-1}\right) \\
\quad=\sum_{i=0}^{n}(n+1) a_{i} x^{i}+\sum_{i=1}^{n} i a_{i} x^{i}+\sum_{i=0}^{n-1}(i+1) a_{i+1} x^{i} \\
\quad=(2 n+1) a_{n} x^{n}+\sum_{i=1}^{n-1}\left((n+1+i) a_{i}+(i+1) a_{i+1}\right) x^{i}+\left((n+1) a_{0}+a_{1}\right)
\end{array}
$$

According to (1), we know:
$$
\left\{\begin{array}{c}
a_{0}=1 \\
(n+1) a_{0}+a_{1}=0 \\
(n+1+i) a_{i}+(i+1) a_{i+1}=0(1 \leq i \leq n-1)
\end{array}\right.
$$

By induction, it is easy to know $a_{i}=(-1)^{i} C_{n+i}^{i}(0 \leq i \leq n)$. Therefore,
$$
Q(x)=\sum_{i=0}^{n}(-1)^{i} C_{n+i}^{i} x^{i}
$$

According to (3), we know $\mathrm{Q}\left(-\frac{1}{2}\right)=\sum_{\mathrm{k}=0}^{\mathrm{n}}(-1)^{\mathrm{k}} \mathrm{C}_{\mathrm{n}+\mathrm{k}}^{\mathrm{k}}\left(-\frac{1}{2}\right)^{\mathrm{k}}=\sum_{\mathrm{k}=0}^{\mathrm{n}} \frac{\mathrm{C}_{n+k}^{\mathrm{n}}}{2^{\mathrm{k}}}$.
$$
\begin{array}{l}
\text { Let } a_{n}=\sum_{k=0}^{n} \frac{C_{n+k}^{n}}{2^{k}}(n \in N) \text {, then } a_{0}=1 \text { and } \\
a_{n+1}=\sum_{k=0}^{n+1} \frac{C_{n+1+k}^{n+1}}{2^{k}}=\sum_{k=0}^{n+1} \frac{C_{n+k}^{n+1}+C_{n+k}^{n}}{2^{k}}=\sum_{k=1}^{n+1} \frac{C_{n+k}^{n+1}}{2^{k}}+\sum_{k=0}^{n} \frac{C_{n+k}^{n}}{2^{k}}+\frac{C_{2 n+1}^{n}}{2^{n+1}} \\
=\sum_{k=0}^{n} \frac{C_{n+k+1}^{n+1}}{2^{k+1}}+\sum_{k=0}^{n} \frac{C_{n+k}^{n}}{2^{k}}+\frac{C_{2 n+2}^{n+1}}{2^{n+2}}=\sum_{k=0}^{n+1} \frac{C_{n+k+1}^{n+1}}{2^{k+1}}+\sum_{k=0}^{n} \frac{C_{n+k}^{n}}{2^{k}} \\
=\frac{a_{n+1}}{2}+a_{n} \\
\end{array}
$$

Rearranging, we get:
$$
\mathrm{a}_{\mathrm{n}+1}=2 \mathrm{a}_{\mathrm{n}}
$$

Thus, we know $Q\left(-\frac{1}{2}\right)=a_{n}=2^{n}$.
Solution method two: According to the condition, we know:
$$
\begin{array}{l}
Q(x)=\frac{1-x^{n+1} P(x)}{(1+x)^{n+1}}=\left(1-x^{n+1} P(x)\right)\left(\sum_{k=0}^{+\infty}\binom{-(n+1)}{k} x^{k}\right) \\
=\left(1-x^{n+1} P(x)\right)\left(\sum_{k=0}^{+\infty}(-1)^{k} C_{n+k}^{k} x^{k}\right) \\
=\sum_{k=0}^{+\infty}(-1)^{k} C_{n+k}^{k} x^{k}-\left(\sum_{k=0}^{+\infty}(-1)^{k} C_{n+k}^{k} x^{n+k+1}\right) P(x)
\end{array}
$$

Noting that $\operatorname{deg} Q \leq n$, comparing the coefficients of each term, we get:
$$
Q(x)=\sum_{k=0}^{\mathrm{n}}(-1)^{\mathrm{k}} C_{n+k}^{\mathrm{k}} \mathrm{x}^{\mathrm{k}}
$$

The rest is the same as method one.
Note: Similarly, we know: $(x-1)^{\mathrm{n}+1} \mathrm{P}(\mathrm{x}-1)+\mathrm{x}^{\mathrm{n}+1} \mathrm{Q}(\mathrm{x}-1)=1$, so
$$
\begin{array}{l}
P(x-1)=\frac{1-x^{n+1} Q(x-1)}{(x-1)^{n+1}}=\left(1-x^{n+1} Q(x-1)\right) \cdot(-1)^{n+1} \cdot\left(\sum_{k=0}^{+\infty} C_{n+k}^{k} x^{k}\right) \\
=(-1)^{n+1} \cdot \sum_{k=0}^{+\infty} C_{n+k}^{k} x^{k}+(-1)^{n}\left(\sum_{k=0}^{+\infty} C_{n+k}^{k} x^{n+k+1}\right) Q(x-1)
\end{array}
$$

Noting that deg $\mathrm{P} \leq \mathrm{n}$, comparing the coefficients of each term, we get:
$$
P(x-1)=(-1)^{n+1} \sum_{k=0}^{n} C_{n+k}^{k} x^{k}
$$

Thus, we know $P(x)=(-1)^{n+1} \sum_{k=0}^{n} C_{n+k}^{k}(x+1)^{k}$.