Let's determine

$$
1 \cdot 4 + 3 \cdot 6 + 5 \cdot 8 + 7 \cdot 10 + \ldots
$$

the $n$-th term of the series, as well as the sum of the first $n$ terms.
The given series can also be written as:

$$
\begin{gathered}
1 \cdot 4 + (1 + 2)(4 + 2) + (1 + 2 \cdot 2)(4 + 2 \cdot 2) + \ldots + \\
+ [1 + (n-1) \cdot 2][4 + (n-1) \cdot 2]
\end{gathered}
$$

Thus, the $n$-th term of the series is

$$
a_{n} = [1 + (n-1) \cdot 2][4 + (n-1) \cdot 2] = 4n^2 + 2n - 2
$$

If we substitute $1, 2, 3, \ldots, n$ for $n$ in this expression, the sum of the series will be:

$$
\begin{aligned}
S = & 4\left(1^2 + 2^2 + 3^2 + \ldots + n^2\right) + 2(1 + 2 + 3 + \ldots + n) - 2n \\
& = 4 \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} - 2n \\
= & \frac{4n^3 + 6n^2 + 2n}{3} + n^2 + n - 2n = \frac{n}{3}(4n^2 + 9n - 1).
\end{aligned}
$$

(Frigyes Riesz.)

The problem was also solved by: Bálint B., Bauss O., Freund A., Friedmann B., Geist E., Goldstein Zs., Goldziher K.; Grünhut B., Hofbauer E., Kántor N., Kiss B., Kornis Ö., Mayer G., Mihalkovics E., Spitzer Ö., Schiffer H., Schwartz E., Szabó K., Thiringer A.