Closed-form expression for the sums of the rows of the Trinomial triangle
Intuitively, each integer on the $i$-th row participates exactly three times in the sum of the integers on the $(i+1)$-th row. This means that the sum $S_{i+1}$ of the integers in the $(i+1)$-th row must be exactly $3$ times as large as the sum $S_i$ of the integers in the $i$-th row, i.e., $S_{i+1}=3S_i$ (and $S_1=1)$.  A formal argument is as follows:
Let $x^{(i)}_j$ be the $j$-th integer on the $i$-th row and define $S_i = \sum_{j=1}^{2i-1} x^{(i)}_j$. For simplicity define $x^{(i)}_j = 0$ whenever $j \le 0$ or $j \ge 2i$. You can show by induction on $i$ that $S_i = 3^{i-1}$.
The base case $i=1$ is trivially true since $S_1 = x^{(1)}_1 = 1 = 3^0 = 3^{1-1}$.
Suppose now that the claim holds for $i\ge 1$. Then, the claim must also hold for $i+1$. Indeed:
$$
\begin{align*}
S_{i+1} &amp;= \sum_{j=1}^{2i+1} x^{(i+1)}_j = \sum_{j=1}^{2i+1} \left( x^{(i)}_{j-2} +  x^{(i)}_{j-1} + x^{(i)}_{j}\right) 
= \sum_{j=1}^{2i+1}  x^{(i)}_{j-2} +  \sum_{j=1}^{2i+1}  x^{(i)}_{j-1} + \sum_{j=1}^{2i+1}  x^{(i)}_{j} \\
&amp;= 3 \sum_{j=1}^{2i-1} x^{(i)}_{j} = 3S_i = 3 \cdot3^{i-1} = 3^i.
\end{align*}
$$