Let's determine the sum of the following series:

$$
3\binom{n}{1}+3^{2}\binom{n}{2}+\ldots+3^{n}\binom{n}{n}
$$
Let's denote the sought sum by $S$, then

$$
S+1=1+\binom{n}{1} 3+\binom{n}{2} 3^{2}+\ldots+\binom{n}{n} 3^{n}=(1+3)^{n}=4^{n}
$$

and thus

$$
S=4^{n}-1
$$

János Fülöp (Bp. V., Piarist g. II. o. t.)