## Task B-4.1.

For real numbers $x$ and $y, x<y$, the following holds:

$$
\begin{aligned}
& x^{3}+y^{3}=65 \\
& 2^{x+y}+2^{x+y+1}+2^{x+y+2}=224
\end{aligned}
$$

Let $y$ be the first, and $x$ the second term of a geometric sequence. What is the sum of this sequence?
## Solution.

From the second given equality, we get: $2^{x+y} \cdot(1+2+4)=224$, which means $x+y=5$.

From the first given equality, we have: $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=65$, which means $x^{2}-x y+y^{2}=13$.

It follows that $x$ and $y$ satisfy the system of equations $x+y=5, x^{2}-x y+y^{2}=13$. By substituting $y=5-x$ into the second equation, we get the quadratic equation $x^{2}-5 x+4=0$, and considering $x<y$, we obtain the solution to the system $x=1, y=4$.

For the geometric series whose first two terms are 4 and 1, we have $q=\frac{1}{4}<1$, so its sum exists and is equal to $S=\frac{4}{1-\frac{1}{4}}=\frac{16}{3}$.

Note: The student could have obtained the solution $x=1, y=4$ in a different way. In this case, obtaining the equality $x+y=5$ is worth 1 point, and setting up and solving the system from which the solution $x=1, y=4$ is obtained is worth 3 points.