As each sum develops it should become clear that the last number in each sum is triangular. So for $S_n$, the last number in the sum is the $n^{th}$ triangular number $= n(n + 1)/2$. Bearing this in mind and the fact that the first number in the sum is the $(n - 1)^{th}$ triangular number plus $1$, then,
Therefore $S_{17} = 17 \times\frac{17^2+1}{2} = 17 \times{290\over2} = 2465$