Example 10 Given that $a, b, x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$

Find the value of $a x^{5}+b y^{5}$.
Given that the expression to be found has a high power, but if we notice that the expression to be found and the known expression share a common feature, that is, the coefficients of $x, y$ in each equation are the same, and the degree of the equation increases by one each time, we can use $x+y$ to multiply each equation to establish a recursive relationship.
From the known conditions, construct the recursive formula $s_{n}=a x^{n}+b y^{n}, n=1,2, \cdots$,
then $s_{n+2}=a x^{n+2}+b y^{n+2}=\left(a x^{n+1}+b y^{n+1}\right)(x+y)-x y\left(a x^{n}+b y^{n}\right)$.
Therefore, $s_{n+2}=(x+y) s_{n+1}-x y s_{n}$.
Substituting the known equations, we get the system of equations
$$
\left\{\begin{array}{l}
16=7(x+y)-3 x y, \\
42=16(x+y)-7 x y .
\end{array}\right.
$$

Solving these, we get $x+y=-14, x y=-38$.
Thus, $s_{5}=(x+y) s_{4}-x y s_{3}=-14 \times 42+38 \times 16=20$.
That is, $a x^{5}+b y^{5}=20$.