Example 5. Let real numbers $a, b, x$ and $y$ satisfy $a x + b y = 3, a x^2 + b y^2 = 7, a x^3 + b y^3 = 16, a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(Eighth American Invitational Mathematics Examination)
Let $S_{\mathrm{n}}=a x^{n}+b y^{\star}, A=x+y, B=x y$, then $x, y$ are the two roots of the equation $u^{2}-A u+B=0$. Therefore,
$$
\begin{array}{l}
x^{2}-A x+B=0, y^{2}-A y+B=0 . \\
\therefore a x^{n}-A\left(a x^{n-1}\right)+B\left(a x^{n-2}\right)=0, \\
b y^{n}-A\left(b y^{n-1}\right)+B b y^{n-2}=0 .
\end{array}
$$

Adding the two equations, we get
$$
\begin{array}{l}
S_{n}=A S_{n-1}-B S_{n-2} . \\
\therefore S_{5}=A S_{4}-B S_{3}=42 A-16 B .
\end{array}
$$

Also, since $S_{4}=A S_{3}-B S_{2}, S_{3}=A S_{2}-B S_{1}$,
$\therefore$ from $S_{1}=3, S_{2}=7, S_{3}=16, S_{4}=42$, we have
$$
42=16 A-7 B, 16=7 A-3 B \text {, }
$$

Solving these, we get $A=-14, B=-38$.
Thus, $S_{5}=42 \times(-14)-16 \times(-38)=20$.