\(a_n = a_{n-1} + 4\) with \(a_1 = 5\text{.}\)
\(a_n = 5 + 4(n-1)\text{.}\)
Yes, since \(2013 = 5 + 4(503-1)\) (so \(a_{503} = 2013\)).
133
\(\frac{538\cdot 133}{2} = 35777\text{.}\)
\(b_n = 1 + \frac{(4n+6)n}{2}\text{.}\)