For all $n \geq 1$, simplify the expression

$$
\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n-1}+\binom{n}{n}
$$
We use the proposition with $a=b=1$: the sum of interest is therefore $(1+1)^{n}=2^{n}$.

Thus, the sum of the coefficients of the $n$-th row of Pascal's triangle is $2^{n}$. We can provide another proof, by induction: the sum of the numbers in the 0-th row of Pascal's triangle is 1, which initializes the induction. Suppose that the sum of the numbers in the $n$-th row is $2^{n}$ and consider the $(n+1)$-th row. Each term in this row is the sum of two terms from the previous row, and conversely, each term in the previous row contributes to two terms in the next row, so the sum of the terms has been multiplied by two, and is $2^{n+1}$, hence the result.