Victor makes a pile with all his marbles so that each marble rests on four marbles from the lower level, the centers of which form a square. This pile has $n$ levels and the last level is a rectangle of $a$ marbles long and $b$ wide. How many marbles does Victor have at a minimum?
A rectangle of size $a \times b$ rests on a minimum rectangle of size $(a+1) \times (b+1)$. Therefore, if Victor's pile contains the fewest possible marbles, the pile resembles a truncated pyramid with a rectangular base. More precisely, the $k$-th level from the top is a rectangle of size $(a+k-1)(b+k-1)$. Thus, in total, Victor has the following number of marbles:

$$
\begin{aligned}
\sum_{k=1}^{n}(a+k-1)(b+k-1) & =\sum_{k=1}^{n} k^{2}+(a+b-2) \sum_{k=1}^{n} k+n(a-1)(b-1) \\
& =\frac{n(n+1)(2 n+1)}{6}+(a+b-2) \frac{n(n+1)}{2}+n(a-1)(b-1) \\
& =n\left((n+1) \frac{2 n+3(a+b)-5}{6}+(a-1)(b-1)\right)
\end{aligned}
$$