What does "the sum of every third element in the $n$-th row of Pascal's triangle" mean? I am looking at the following problem. I don't want to know how it's done,  I would just like to see the problem reworded in less confusing terms if possible:

 A: $S_{n,0}$ is the sum of every third element in row $n$ of Pascal's triangle, starting with the first element in that row.  So, for example, $S_{5,0}$ is the sum of every third element in row 5 of Pascal's triangle, starting with the first element in row 5.  Then we get $S_{5,0} = 1 + 10 = 11$.  The elements we're using in row 5 are in red below: $$ \color{red}{1} \quad 5 \quad 10 \quad \color{red}{10} \quad 5 \quad 1 $$
$S_{n,1}$ is the sum of every third element in row $n$ of Pascal's triangle, starting with the second element in that row.  So, for example, $S_{5,1}$ is the sum of every third element in row 5 of Pascal's triangle, starting with the second element in row 5.  Then we get $S_{5,1} = 5 + 5 = 10$.  The elements we're using in row 5 are in red below: $$ 1 \quad \color{red}{5} \quad 10 \quad 10 \quad \color{red}{5} \quad 1 $$
$S_{n,2}$ is the sum of every third element in row $n$ of Pascal's triangle, starting with the third element in that row.  So, for example, $S_{5,2}$ is the sum of every third element in row 5 of Pascal's triangle, starting with the third element in row 5.  Then we get $S_{5,2} = 10 + 1 = 11$.  The elements we're using in row 5 are in red below: $$ 1 \quad 5 \quad \color{red}{10} \quad 10 \quad 5 \quad \color{red}{1} $$
Make a conjecture about the value of $S_{100,1}$.  That is, see if you can determine a pattern that will allow you to calculate the value of $S_{100,1}$, which is the sum of every third element in row $100$ of Pascal's triangle, starting with the 2nd element in that row.
A: Let ${n\choose k}$ be the $k^{th}$ entry of the $n^{th}$ row of the triangle.  (starting at $k=0$)
$S_{n,0} = {n\choose 0}+{n\choose 3}+{n\choose 6}+\cdots\\
S_{n,1} = {n\choose 1}+{n\choose 4}+{n\choose 7}+\cdots\\
S_{n,2} = {n\choose 2}+{n\choose 5}+{n\choose 8}+\cdots$
