Note that: \begin{align*} a_1 = \frac{1 \times 2}{2} &= \frac {1}{6} (1\times 2 \times 3 - 0 \times 1 \times 2) \\ a_2 = \frac{2 \times 3}{2} &= \frac {1}{6} (2\times 3 \times 4 - 1\times 2 \times 3) \\ a_3 = \frac{3 \times 4}{2} &= \frac {1}{6} (3\times 4 \times 5 - 2\times 3 \times 4) \\ \\ a_n = \frac{n (n + 1)}{2} &= \frac {1}{6} [n (n+1) (n+2) - (n-1) n (n+1)] \end{align*}$$b_n=\sum_{i=1}^n a_i = \sum_{i=1}^n \frac {i(i+1)}{2}=\frac {1}{6} n (n+1) (n+2)$$$$6b_n = (n+1)^3-(n+1)$$Observe the units digit of cube, we have: $$n \rightarrow n^3 \Rightarrow (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) \rightarrow (0, 1, 8, 7, 4, 5, 6, 3, 2, 9)$$The units digit of $(n+1)^3-(n+1)$ cannot be $2$, then units digit of $b_n$ cannot be $7$, and thus $b_n$ cannot be $2017$.