Sebestyén, starting from his third birthday, receives a triangular prism-shaped cake every birthday, with a candle at each of the top three vertices, and on top, as many candles as his age, so that the total number of candles on the cake equals his age, with no three candles lying on the same straight line. Sebestyén likes to cut the cake into triangular slices, the vertices of which are given by the positions of the candles (the interior of the triangles must not contain any candles). How many slices can the cake be divided into on his $k$-th birthday?
Precisely $2k-5$ slices can be obtained. Let's prove this by complete induction. We can start with the case $k=3$, in which it is self-evident that the cake can be divided into exactly one slice. In the case of $k=4$, it is also immediately apparent that the cake can be divided into 3 slices.

Assume that we have proven that up to $k$, the cake can be divided into exactly $2k-5$ slices every year. Let's prove that then on the $(k+1)$-th birthday, the cake can be divided into exactly $2(k+1)-5$ slices.

Take any $(k+1)$-candle triangular cake. Consider any valid slicing of this, and then select any interior candle (i.e., any that is not at the apex of the cake). Let the number of triangle sides (i.e., the number of cuts) originating from this be $n$. $n$ is at least 3, since no three candles are collinear and the cake surface is divided into triangles. Thus, there is exactly one $n$-gon that the slicing defines and contains only the selected candle. The candles at the ends of the cuts originating from the selected candle are connected to each other, otherwise either the cake would not be sliced into triangles, or there would be a cut from the selected candle that we missed.

The $n$ adjacent candles define an $n$-gon, inside which only the selected candle is located. This candle is connected to every vertex, so the $n$-gon is divided into $n$ triangles. Now remove the selected candle. This leaves a $k$-candle cake, which, by the induction hypothesis, can be divided into exactly $2k-5$ slices. Do not change the slicing outside the $n$-gon. Choose one vertex of the $n$-gon, and connect it to all other non-adjacent vertices. This divides the $n$-gon into $n-2$ triangles. It is not possible to divide the $n$-gon into a different number of triangles, since in any such division, each triangle's angles are part of the $n$-gon's angles, so the sum of the $180^\circ$ angles of the $t$ triangles is the $180(n-2)$ degrees sum of the $n$-gon's interior angles, from which $t=n-2$ is uniquely determined.

Therefore, when there were $k+1$ candles, the cake could be divided into 2 more slices, totaling $2k-5+2=2(k+1)-5$ slices.

This proves that on Sebestyén's $k$-th birthday, the cake can only be divided into $2k-5$ slices.