What comes after 1,2,3,4?

If $a + b = 1$ , then $b = 1 - a$ .

If $ax + by = 2$ and $b = 1 - a$ , then $y = \frac{2 - ax}{1 - a}$ .

If $ax^2 + by^2 = 3$ and $b = 1 - a$ , then $y^2 = \frac{3 - ax^2}{1 - a}$ , and since $y = \frac{2 - ax}{1 - a}$ , we have $(\frac{2 - ax}{1 - a})^2 = \frac{3 - ax^2}{1 - a}$ , which solves to $a = \frac{-1}{(x - 1)(x - 3)}$ .

If $ax^3 + by^3 = 4$ and $b = 1 - a$ , then $y^3 = \frac{4 - ax^3}{1 - a}$ , and since $y = \frac{2 - ax}{1 - a}$ , we have $(\frac{2 - ax}{1 - a})^3 = \frac{4 - ax^3}{1 - a}$ , and since $a = \frac{-1}{(x - 1)(x - 3)}$ , we have $\Big(\frac{2 + \frac{x}{(x - 1)(x - 3)}}{1 + \frac{1}{(x - 1)(x - 3)}}\Big)^3 = \frac{4 + \frac{x^3}{(x - 1)(x - 3)}}{1 + \frac{1}{(x - 1)(x - 3)}}$ , which solves to $x = 2$ .

But if $x = 2$ , then $a = \frac{-1}{(x - 1)(x - 3)} = \frac{-1}{(2 - 1)(2 - 3)} = 1$ , which is impossible since $y = \frac{2 - ax}{1 - a} = \frac{2 - 1 \cdot 1}{1 - 1} = \frac{0}{0}$ . Therefore, there are no possible values for $a$ , $b$ , $x$ , and $y$ under the given conditions.

If there were $a,b,x,y$ such that $a+b=1 \hspace{1cm} ax + by = 2 \hspace{1cm} ax^2 + by^2 = 3 \hspace{1cm} ax^3 + by^3 = 4$ then let $u = x+y$ , $v = xy$ . Then $x,y$ are the roots of the quadratic equation $f(X) \; = \; X^2 - uX + v \; = \; 0$ But then $x^{n+2} - ux^{n+1} + vx^n \; = \; y^{n+2} - uy^{n+1} + vy^n \; = \; 0 \hspace{2cm} n \in \mathbb{N} \cup \{0\}$ and hence $A_{n+2} - uA_{n+1} + vA_n \; = \; 0 \hspace{2cm} n \in \mathbb{N} \cup \{0\}$ where $A_n \; = \; ax^n + by^n$ But this would imply that $3 - 2u + v = 4 - 3u + 2v = 0$ , and hence $u=2,v=1$ , so that $f(X) = X^2 - 2X + 1$ . But this means that $x=y=1$ , which in turn implies that $a_n = a+b = 1$ for all $n \ge 1$ .

Thus no complex numbers $a,b,x,y$ exist with the desired properties. This means that none of the others is the best possible answer, although this is not possible would have been a better answer.