Let's write the polynomial

$$
S=x+3 x^{2}+6 x^{3}+10 x^{4}+\ldots+\frac{n(n+1)}{2} x^{n}
$$

in a simpler form, if $x \neq 1$.
Let's notice that in the consecutive terms of $S$, the coefficients are the sums of the first $1, 2, 3, 4, \ldots, n$ natural numbers:

$$
3=1+2, \quad 6=1+2+3, \quad \ldots, \quad n(n+1) / 2=1+2+\ldots+n
$$

Accordingly, $S$ can be written as:

$$
\begin{aligned}
S & =x+ \\
& +x^{2}+2 x^{2}+ \\
& +x^{3}+2 x^{3}+3 x^{3}+ \\
& +x^{4}+2 x^{4}+3 x^{4}+4 x^{4}+ \\
& +\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+ \\
& +x^{n}+2 x^{n}+3 x^{n}+4 x^{n}+\ldots+n x^{n}
\end{aligned}
$$

Here, the number of columns is $n$, and each column is a geometric series with $n, n-1, n-2, \ldots, 1$ terms, respectively. The common ratio in each series is $x \neq 1$, so we can apply the formula for the sum of a geometric series:

$$
\begin{aligned}
S & =\frac{x\left(x^{n}-1\right)}{x-1}+\frac{2 x^{2}\left(x^{n-1}-1\right)}{x-1}+\frac{3 x^{3}\left(x^{n-2}-1\right)}{x-1}+\ldots+\frac{n x^{n}(x-1)}{x-1}= \\
& =\frac{x}{x-1}\left[\left(x^{n}+2 x^{n}-3 x^{n}+\ldots n x^{n}\right)-\left(1+2 x+3 x^{2}+\ldots+n x^{n-1}\right)\right]
\end{aligned}
$$

The first part in the brackets is $n(n+1) x^{n} / 2$, and the subtrahend $S^{\prime}$ can be similarly transformed:

$$
\begin{gathered}
S^{\prime}=1+ \\
+x+x+ \\
+x^{2}+x^{2}+x^{2}+ \\
+\ldots \ldots \ldots \ldots \ldots+ \\
+x^{n-1}+x^{n-1}+x^{n-1}+\ldots+x^{n-1}= \\
=\frac{x^{n}-1}{x-1}+\frac{x\left(x^{n-1}-1\right)}{x-1}+\ldots+\frac{x^{n-1}(x-1)}{x-1}= \\
=\frac{1}{x-1}\left[n x^{n}-\left(1+x+x^{2}+\ldots+x^{n-1}\right)\right]=\frac{1}{x-1}\left[n x^{n}-\frac{x^{n}-1}{x-1}\right]= \\
=\frac{n x^{n}}{x-1}-\frac{\left(x^{n}-1\right)}{(x-1)^{2}}
\end{gathered}
$$

Thus, the original expression is:

$$
S=\frac{n(n+1) x^{n+1}}{2(x-1)}-\frac{n x^{n+1}}{(x-1)^{2}}+\frac{x\left(x^{n}-1\right)}{(x-1)^{3}}
$$

Halász Ákos (Kecskemét, Piarist g. IV. class)