We are all familiar with Pascal's triangle.
1
11
121
1331
14641
...
Conveniently, each row is actually all the coefficients of $n \choose k$ for row $n$. So row 0, is $0 \choose 0$, row 1 is $1 \choose 0$, $1 \choose 1$, row 2 is $2 \choose 0$, $2 \choose 1$, $2 \choose 2$ and so on...
Note that at any row $n$ we have exactly $n+1$ coefficients, I am interested in the sum $\sum^k_{i=0} {n \choose i}$ or in English, the sum of the first k coefficients of the row.
Is there a direct formula that allows you to compute the sum if you know $n, k$? i.e. something similar to $\frac{n^2 + n}{2}$ for the sum of the naturals but for pascals triangle, that likewise gives a direct result without needing to add each term individually in a for loop.
