Given that the first column and diagonal of a triangle are defined, the terms in the rows can be created in ways other than how it is done for Pascal's triangle, for which the sum of the terms is easily calculated. The alternatives are $T(i,j)=T(i,j-1) + T(i-1,j-1)$ or $T(i,j)=T(i,j-1) + T(i-1,j).$
Is there a simple way to find the sum of the terms $i$ in each row for these triangles? So far all I've found is a rather messy summation growing in complexity as the rows increase.
One could experiment with triangles that give different row sums when the arms are exchanged one for the other and symmetry does not hold. Of course, one could not care to experiment to uncover oddities.
The $n$th row of Pascal's Triangle contains, reading left to right $$\binom{n}{0},\binom{n}{1},\binom{n}{2},\cdots,\binom{n}{n-1},\binom{n}{n}$$ Where $$\binom{n}{r}=\frac{n!}{r!(n-r)!}$$
Hence the sum you're looking for is: $$ \sum_{r=0}^{n}{\binom{n}{r}}$$
Which evaluates to $2^n$. Hence the sum of the $n$th row of Pascal's Triangle is $2^n$