1. Define the function $f(k) = ax^k + by^k$. Given the equations:
$$\begin{align*} f(2014) &= 6, \\ f(2015) &= 7, \\ f(2016) &= 3, \\ f(2017) &= 50, \end{align*}$$
we need to find $f(2018)$.

2. Observe that $f(k)$ follows a linear recurrence relation. We can derive this by considering the expressions for $f(k+1)$ and $f(k-1)$:
$$f(k+1) = ax^{k+1} + by^{k+1} = x(ax^k + by^k) + y(ax^k + by^k) - xy(ax^{k-1} + by^{k-1}).$$
Simplifying, we get:
$$f(k+1) = (x+y)f(k) - xyf(k-1).$$
Let $p = x + y$ and $q = -xy$. Then the recurrence relation becomes:
$$f(k+1) = pf(k) + qf(k-1).$$

3. Substitute the known values into the recurrence relation to find $p$ and $q$:
$$\begin{align*} f(2016) &= pf(2015) + qf(2014), \\ 3 &= 7p + 6q, \\ f(2017) &= pf(2016) + qf(2015), \\ 50 &= 3p + 7q. \end{align*}$$

4. Solve the system of linear equations:
$$\begin{cases} 3 = 7p + 6q, \\ 50 = 3p + 7q. \end{cases}$$
Multiply the first equation by 7 and the second by 6 to eliminate $q$:
$$\begin{align*} 21 &= 49p + 42q, \\ 300 &= 18p + 42q. \end{align*}$$
Subtract the second equation from the first:
$$21 - 300 = 49p - 18p \implies -279 = 31p \implies p = -9.$$
Substitute $p = -9$ back into one of the original equations:
$$3 = 7(-9) + 6q \implies 3 = -63 + 6q \implies 66 = 6q \implies q = 11.$$

5. The recurrence relation is now:
$$f(k+1) = -9f(k) + 11f(k-1).$$

6. Use the recurrence relation to find $f(2018)$:
$$\begin{align*} f(2018) &= -9f(2017) + 11f(2016), \\ &= -9(50) + 11(3), \\ &= -450 + 33, \\ &= -417. \end{align*}$$