Initial 52. Given real numbers $a, b, x, y$ satisfy $a x^{n}+b y^{n}=1+2^{n+1}$ for any natural number $n$. Find the value of $x^{a}+y^{b}$.
Let $S_{n}=a x^{\prime \prime}+b y^{\prime \prime}$. Then
$$
S_{1}=a x+b y=5 \text {, }
$$
$$
\begin{array}{l}
S_{0}=a x^{2}+b y^{2}=9 . \\
S=a x^{3}+b y^{3}=17 . \\
S_{1}=a x^{4}+b y^{\prime}=33 .
\end{array}
$$

Let $x+y=A, x y=B$. Then $x, y$ are the roots of the quadratic equation $t^{2}-A t+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, \\
y^{2}-A y+B=0 .
\end{array}
$$
a. $x^{n-2} \cdot$ (3) $+b y^{n-2} \cdot$ (1), we get
$$
S_{n}=A S_{n-1}-B S_{n-2} \text {. }
$$

From the problem, we have $\left\{\begin{array}{l}S_{3}=A S_{2}-B S_{1}, \\ S_{4}=A S_{3}-B S_{2},\end{array}\right.$
which gives $\left\{\begin{array}{l}17=9 A-5 B, \\ 33=17 A-9 B .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}A=3, \\ B=2 .\end{array}\right.$.
Thus, $\left\{\begin{array}{l}x+y=3 \\ x y=2\end{array}\right.$

Substituting $x=1, y=2$ into (1) and (2), we get $a=1, b=2$.
Similarly, when $x=2, y=1$, we get $a=2, b=1$.
(1) When $x=1, y=2, a=1, b=2$,
$$
x^{4}+y^{b}=1^{1}+2^{2}=5 \text {. }
$$
(2) When $x=2, y=1, a=2, b=1$,
$$
x^{2}+y^{3}=1^{1}+2^{2}=5 \text {. }
$$
（From Yu Yuanzheng, "Zhonghua Yinghao School, Conghua, Guangzhou, 510960）